If $a + 1/b = b + 1/c = c + 1/a$, how to find the value of $abc$?












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If $a, b, c$ be distinct reals such that $$a + frac1b = b + frac1c = c + frac1a ,$$ how do I find the value of $abc$?




The answer says $1$, but I am not sure how to derive it.












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  • $begingroup$
    What have you tried? By the way, do you see a simple particular solution for $a$, $b$, $c$?
    $endgroup$
    – Shaun Ault
    Aug 31 '11 at 22:34






  • 1




    $begingroup$
    There is a large amount of hidden symmetry in this problem (well beyond the cyclic substitution that takes A to B to C to A). Where did the question come from? Any reference?
    $endgroup$
    – zyx
    Sep 5 '11 at 6:42










  • $begingroup$
    @zyx Interestingly, this was asked in a $2017$ Olympiad.
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 16 '18 at 17:55
















13












$begingroup$



If $a, b, c$ be distinct reals such that $$a + frac1b = b + frac1c = c + frac1a ,$$ how do I find the value of $abc$?




The answer says $1$, but I am not sure how to derive it.












share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried? By the way, do you see a simple particular solution for $a$, $b$, $c$?
    $endgroup$
    – Shaun Ault
    Aug 31 '11 at 22:34






  • 1




    $begingroup$
    There is a large amount of hidden symmetry in this problem (well beyond the cyclic substitution that takes A to B to C to A). Where did the question come from? Any reference?
    $endgroup$
    – zyx
    Sep 5 '11 at 6:42










  • $begingroup$
    @zyx Interestingly, this was asked in a $2017$ Olympiad.
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 16 '18 at 17:55














13












13








13


10



$begingroup$



If $a, b, c$ be distinct reals such that $$a + frac1b = b + frac1c = c + frac1a ,$$ how do I find the value of $abc$?




The answer says $1$, but I am not sure how to derive it.












share|cite|improve this question











$endgroup$





If $a, b, c$ be distinct reals such that $$a + frac1b = b + frac1c = c + frac1a ,$$ how do I find the value of $abc$?




The answer says $1$, but I am not sure how to derive it.









algebra-precalculus






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edited Dec 16 '14 at 0:06









Mike Miller

37.6k473140




37.6k473140










asked Aug 31 '11 at 22:16









QuixoticQuixotic

11.7k2388178




11.7k2388178












  • $begingroup$
    What have you tried? By the way, do you see a simple particular solution for $a$, $b$, $c$?
    $endgroup$
    – Shaun Ault
    Aug 31 '11 at 22:34






  • 1




    $begingroup$
    There is a large amount of hidden symmetry in this problem (well beyond the cyclic substitution that takes A to B to C to A). Where did the question come from? Any reference?
    $endgroup$
    – zyx
    Sep 5 '11 at 6:42










  • $begingroup$
    @zyx Interestingly, this was asked in a $2017$ Olympiad.
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 16 '18 at 17:55


















  • $begingroup$
    What have you tried? By the way, do you see a simple particular solution for $a$, $b$, $c$?
    $endgroup$
    – Shaun Ault
    Aug 31 '11 at 22:34






  • 1




    $begingroup$
    There is a large amount of hidden symmetry in this problem (well beyond the cyclic substitution that takes A to B to C to A). Where did the question come from? Any reference?
    $endgroup$
    – zyx
    Sep 5 '11 at 6:42










  • $begingroup$
    @zyx Interestingly, this was asked in a $2017$ Olympiad.
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 16 '18 at 17:55
















$begingroup$
What have you tried? By the way, do you see a simple particular solution for $a$, $b$, $c$?
$endgroup$
– Shaun Ault
Aug 31 '11 at 22:34




$begingroup$
What have you tried? By the way, do you see a simple particular solution for $a$, $b$, $c$?
$endgroup$
– Shaun Ault
Aug 31 '11 at 22:34




1




1




$begingroup$
There is a large amount of hidden symmetry in this problem (well beyond the cyclic substitution that takes A to B to C to A). Where did the question come from? Any reference?
$endgroup$
– zyx
Sep 5 '11 at 6:42




$begingroup$
There is a large amount of hidden symmetry in this problem (well beyond the cyclic substitution that takes A to B to C to A). Where did the question come from? Any reference?
$endgroup$
– zyx
Sep 5 '11 at 6:42












$begingroup$
@zyx Interestingly, this was asked in a $2017$ Olympiad.
$endgroup$
– Mohammad Zuhair Khan
Dec 16 '18 at 17:55




$begingroup$
@zyx Interestingly, this was asked in a $2017$ Olympiad.
$endgroup$
– Mohammad Zuhair Khan
Dec 16 '18 at 17:55










5 Answers
5






active

oldest

votes


















15












$begingroup$

From the fact that Expressions $1$ and $2$ are equal, we obtain
$$a-b=frac{1}{c}-frac{1}{b}=frac{b-c}{bc}.$$
From the fact that Expressions $2$ and $3$ are equal, we obtain
$$b-c=frac{1}{a}-frac{1}{c}=frac{c-a}{ca}.$$
From the fact that Expressions $3$ and $1$ are equal, we obtain
$$c-a=frac{1}{b}-frac{1}{a}=frac{a-b}{ab}.$$



Multiply the left-hand sides, the right-hand sides. We get
$$(a-b)(b-c)(c-a)=frac{(b-c)(c-a)(a-b)}{(abc)^2}.$$
Since $a$, $b$, and $c$ are distinct, $(a-b)(b-c)(c-a)ne 0$. We conclude that
$(abc)^2=1$. This yields the two possibilities $abc=1$ and $abc=-1$.



In a logical sense we are finished: We have shown that if $(a,b,c)$ is a solution of the system with $a$, $b$, and $c$ distinct, then $abc=pm 1$.



But it would be nice to show that there are solutions of the desired type. So let's find such a solution, with $abc=1$.



Look for a solution with $a=1$. Then we need $c=1/b$. In order to satisfy our equations, we need $1+1/b=2b$. Beside the useless solution $b=1$, this has the solution $b=-1/2$. We conclude that
$$a=1,quad b=-frac{1}{2},quad c=-2$$
is a solution of the desired type, with $abc=1$. By changing all the signs, we find that
$$a=-1,quad b=frac{1}{2},quad c=2$$
is a solution with $abc=-1$.



Added: It is easy to see that if two 0f $a$, $b$, $c$ are equal, then they are all equal, giving the parametric family $(t,t,t)$, where $tne 0$. Now assume that $abc=pm 1$. We find all solutions with $abc=1$. The solutions with $abc=-1$ are then obtained by changing all the signs.



Let $a=t$. Our equations will be satisfied if $t+1/b=b+1/c$, where $c=1/bt$. We therefore obtain the equation $t+1/b=b+bt$, which simplifies to $(1+t)b^2 -tb-1=0$. Now we can solve this quadratic equation for $b$, and get $c$ from $tbc=1$. There is undoubtedly a more symmetric way to obtain the complete parametric description of the solutions!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This method is exactly the same as that presented in Aryabhata's answer, except for the choice of notation: your $: a-b = (b-c)/(bc)$ vs. his $:(a-b):A = a:(b-c):$ for $A = abc:.:$ But the key point is to exploit the innate symmetry, which is more clearly presented above.
    $endgroup$
    – Bill Dubuque
    Sep 1 '11 at 2:20








  • 1




    $begingroup$
    @Bill Dubuque: After I had posted the solution, I noticed that Aryabhata must have posted a solution while I was typing. Wondered whether to delete, decided not to because the line of attack that I used, though essentially identical, is probably more natural from a student point of view. And yes, I did want to pound in the use of symmetry!
    $endgroup$
    – André Nicolas
    Sep 1 '11 at 2:56



















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Give the common value of $a+1/b$ etc. a name, say $h$. We can now rewrite $a+1/b=h$ to $a=h-1/b$ and similarly $b=h-1/c$ and $c=h-1/a$. Telescoping these expressions into each other gives
$$a=frac{h^2-ah-1}{h^3-ah^2-2h+a}$$
which rearranges to
$$(1-h^2)a^2 + (h^3-h)a + 1-h^2 = 0$$
Now because everything is symmetric, $b$ and $c$ must satisfy the same equation, but $a$, $b$ and $c$ were distinct numbers, so they can only be roots in a quadratic polynomial if it is identically zero. So $h^2=1$, which makes all of the coefficients vanish. Thus $h=pm 1$.



Assuming that $h=1$ we now get $b=frac{1}{1-a}$ and $c=1-frac{1}{a}=frac{a-1}{a}$, hence $abc=-1$. An example is $(a,b,c)=(2,-1,1/2)$.



Negating everything in the $h=1$ case gives $h=-1$ and $abc=1$.






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    12












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    Let $A = abc$. We can assume $A ne 0$.



    Now the equations can be written as



    $a + ac/A = b + ab/A = c + bc/A$



    Multiply by $A$ throughout.



    $a(A+c) = b(A+a) = c(A+b)$



    Subtract first two, last two, first and last to get three equations and multiply those to get an equation in $A$.






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      7












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      Below I explicitly highlight the symmetry implicit in the answers of Aryabhata and Andre. Let $rm:sigma:$ be the permutation $rm:(a b c):$ acting on polynomials in $rm:a,b,c:$ by cyclically permuting the variables, i.e. $rm:sigma f(a,b,c): =: f(b,c,a):, $ e.g. $rm sigma(a-b) = b-c:.:$ Define the norm by $rm:N(g) = g sigma g sigma^2 g:.:$ Note that $rm:N(sigma: g): =: N(g):$ since $rm:sigma^3 = 1:.:$ Put $rm:f: =: abc: =: N(a):,:$ so $rm:sigma: f: =: f:.:$ Put $rm:g: =: a-b:.:$ Subtracting each pair of the given three equations yields the following equations



      $$begin{eqnarray*} rm f &=&rm a dfrac{b-c}{a-b} &=&rm a dfrac{sigma: g}{g}



      rm f &=&rm b dfrac{c-a}{b-c} &=&rm :sigma a dfrac{sigma^2 g}{sigma: g}



      rm f &=&rm c dfrac{a-b}{c-a} &=&rm sigma^2 a dfrac{sigma^3:g}{sigma^2g}
      end{eqnarray*}$$



      Multiplying the above, using $rm:sigma^3:g = g:,:$ yields $rm:f^{:3} = N(a) = f:,:$ so $rm:f ne 0:$ $:Rightarrow:$ $rm:f^{:2} = 1:.:$ That's essentially the method employed in said answers. I show that this method amounts to simply taking the norm of the first equation. Notice that the second and third equations result from successively applying $rm:sigma:$ to the first equation, i.e. the given three equations are equivalent to stating that the first equation is preserved by $rm:sigma:$ and $rm:sigma^2:$. So multiplying them all amounts to taking the norm of the first, i.e.



      $$rm f^{:3} = N(f) = Nbigg(a:dfrac{sigma g}{g}bigg) = N(a) = f quad by quad N(sigma g) = N(g) $$



      So, from this viewpoint, the proof that $rm:f^{:3} = f:$ is a one-line inference achieved by by taking a norm (that $rm:f/a = sigma:g/g:$ has norm $1$ is essentially multiplicative telescopy). This simplicity above results from recognizing and exploiting the innate symmetry in the problem.



      Less trivial exploitation of similar symmetries arise in the Galois theory of difference equations and radical extensions (Kummer theory). For one simple example see this answer.






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      • $begingroup$
        Nice answer, bringing out the structure explicitly.
        $endgroup$
        – André Nicolas
        Sep 1 '11 at 4:53



















      0












      $begingroup$

      I'm not sure how different this is, but here is my version



      $a + frac{1}{b} = b + frac{1}{c} = c + frac{1}{a}
      quad$ (Note this implies $abc ne 0$)



      $a^2bc + ac = ab^2c + ab = abc^2 + bc$



      $a(abc) + ac = b(abc) + ab = c(abc) + bc$



      $quad a(abc) + ac = b(abc) + ab implies (a-b)abc=a(b-c)$

      $$ a=b=c ne 0 ; text{ or } ; abc = dfrac{a(b-c)}{a-b}$$



      $quad a(abc) + ac = c(abc) + bc implies (a-c)abc = c(b-a)$
      $$a=c=b ne 0 ; text{ or } ; abc = dfrac{c(b-a)}{a-c}$$



      $quad b(abc) + ab = c(abc) + bc implies (b-c)abc=b(c-a)$
      $$b=c=a ne 0 ; text{ or } ; abc = dfrac{b(c-a)}{b-c}$$



      So, one solution is $; a=b=c ne 0$.

      But if $a,b,$ and $c$ are distinct and non zero, then



      $(abc)^3 =
      dfrac{a(b-c)}{a-b} cdot dfrac{c(b-a)}{a-c} cdot dfrac{b(c-a)}{b-c}$



      $(abc)^3 = abc$



      $(abc)^2 = 1$



      $abc = pm 1$



      $|abc| = 1$






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        5 Answers
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        5 Answers
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        15












        $begingroup$

        From the fact that Expressions $1$ and $2$ are equal, we obtain
        $$a-b=frac{1}{c}-frac{1}{b}=frac{b-c}{bc}.$$
        From the fact that Expressions $2$ and $3$ are equal, we obtain
        $$b-c=frac{1}{a}-frac{1}{c}=frac{c-a}{ca}.$$
        From the fact that Expressions $3$ and $1$ are equal, we obtain
        $$c-a=frac{1}{b}-frac{1}{a}=frac{a-b}{ab}.$$



        Multiply the left-hand sides, the right-hand sides. We get
        $$(a-b)(b-c)(c-a)=frac{(b-c)(c-a)(a-b)}{(abc)^2}.$$
        Since $a$, $b$, and $c$ are distinct, $(a-b)(b-c)(c-a)ne 0$. We conclude that
        $(abc)^2=1$. This yields the two possibilities $abc=1$ and $abc=-1$.



        In a logical sense we are finished: We have shown that if $(a,b,c)$ is a solution of the system with $a$, $b$, and $c$ distinct, then $abc=pm 1$.



        But it would be nice to show that there are solutions of the desired type. So let's find such a solution, with $abc=1$.



        Look for a solution with $a=1$. Then we need $c=1/b$. In order to satisfy our equations, we need $1+1/b=2b$. Beside the useless solution $b=1$, this has the solution $b=-1/2$. We conclude that
        $$a=1,quad b=-frac{1}{2},quad c=-2$$
        is a solution of the desired type, with $abc=1$. By changing all the signs, we find that
        $$a=-1,quad b=frac{1}{2},quad c=2$$
        is a solution with $abc=-1$.



        Added: It is easy to see that if two 0f $a$, $b$, $c$ are equal, then they are all equal, giving the parametric family $(t,t,t)$, where $tne 0$. Now assume that $abc=pm 1$. We find all solutions with $abc=1$. The solutions with $abc=-1$ are then obtained by changing all the signs.



        Let $a=t$. Our equations will be satisfied if $t+1/b=b+1/c$, where $c=1/bt$. We therefore obtain the equation $t+1/b=b+bt$, which simplifies to $(1+t)b^2 -tb-1=0$. Now we can solve this quadratic equation for $b$, and get $c$ from $tbc=1$. There is undoubtedly a more symmetric way to obtain the complete parametric description of the solutions!






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          This method is exactly the same as that presented in Aryabhata's answer, except for the choice of notation: your $: a-b = (b-c)/(bc)$ vs. his $:(a-b):A = a:(b-c):$ for $A = abc:.:$ But the key point is to exploit the innate symmetry, which is more clearly presented above.
          $endgroup$
          – Bill Dubuque
          Sep 1 '11 at 2:20








        • 1




          $begingroup$
          @Bill Dubuque: After I had posted the solution, I noticed that Aryabhata must have posted a solution while I was typing. Wondered whether to delete, decided not to because the line of attack that I used, though essentially identical, is probably more natural from a student point of view. And yes, I did want to pound in the use of symmetry!
          $endgroup$
          – André Nicolas
          Sep 1 '11 at 2:56
















        15












        $begingroup$

        From the fact that Expressions $1$ and $2$ are equal, we obtain
        $$a-b=frac{1}{c}-frac{1}{b}=frac{b-c}{bc}.$$
        From the fact that Expressions $2$ and $3$ are equal, we obtain
        $$b-c=frac{1}{a}-frac{1}{c}=frac{c-a}{ca}.$$
        From the fact that Expressions $3$ and $1$ are equal, we obtain
        $$c-a=frac{1}{b}-frac{1}{a}=frac{a-b}{ab}.$$



        Multiply the left-hand sides, the right-hand sides. We get
        $$(a-b)(b-c)(c-a)=frac{(b-c)(c-a)(a-b)}{(abc)^2}.$$
        Since $a$, $b$, and $c$ are distinct, $(a-b)(b-c)(c-a)ne 0$. We conclude that
        $(abc)^2=1$. This yields the two possibilities $abc=1$ and $abc=-1$.



        In a logical sense we are finished: We have shown that if $(a,b,c)$ is a solution of the system with $a$, $b$, and $c$ distinct, then $abc=pm 1$.



        But it would be nice to show that there are solutions of the desired type. So let's find such a solution, with $abc=1$.



        Look for a solution with $a=1$. Then we need $c=1/b$. In order to satisfy our equations, we need $1+1/b=2b$. Beside the useless solution $b=1$, this has the solution $b=-1/2$. We conclude that
        $$a=1,quad b=-frac{1}{2},quad c=-2$$
        is a solution of the desired type, with $abc=1$. By changing all the signs, we find that
        $$a=-1,quad b=frac{1}{2},quad c=2$$
        is a solution with $abc=-1$.



        Added: It is easy to see that if two 0f $a$, $b$, $c$ are equal, then they are all equal, giving the parametric family $(t,t,t)$, where $tne 0$. Now assume that $abc=pm 1$. We find all solutions with $abc=1$. The solutions with $abc=-1$ are then obtained by changing all the signs.



        Let $a=t$. Our equations will be satisfied if $t+1/b=b+1/c$, where $c=1/bt$. We therefore obtain the equation $t+1/b=b+bt$, which simplifies to $(1+t)b^2 -tb-1=0$. Now we can solve this quadratic equation for $b$, and get $c$ from $tbc=1$. There is undoubtedly a more symmetric way to obtain the complete parametric description of the solutions!






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          This method is exactly the same as that presented in Aryabhata's answer, except for the choice of notation: your $: a-b = (b-c)/(bc)$ vs. his $:(a-b):A = a:(b-c):$ for $A = abc:.:$ But the key point is to exploit the innate symmetry, which is more clearly presented above.
          $endgroup$
          – Bill Dubuque
          Sep 1 '11 at 2:20








        • 1




          $begingroup$
          @Bill Dubuque: After I had posted the solution, I noticed that Aryabhata must have posted a solution while I was typing. Wondered whether to delete, decided not to because the line of attack that I used, though essentially identical, is probably more natural from a student point of view. And yes, I did want to pound in the use of symmetry!
          $endgroup$
          – André Nicolas
          Sep 1 '11 at 2:56














        15












        15








        15





        $begingroup$

        From the fact that Expressions $1$ and $2$ are equal, we obtain
        $$a-b=frac{1}{c}-frac{1}{b}=frac{b-c}{bc}.$$
        From the fact that Expressions $2$ and $3$ are equal, we obtain
        $$b-c=frac{1}{a}-frac{1}{c}=frac{c-a}{ca}.$$
        From the fact that Expressions $3$ and $1$ are equal, we obtain
        $$c-a=frac{1}{b}-frac{1}{a}=frac{a-b}{ab}.$$



        Multiply the left-hand sides, the right-hand sides. We get
        $$(a-b)(b-c)(c-a)=frac{(b-c)(c-a)(a-b)}{(abc)^2}.$$
        Since $a$, $b$, and $c$ are distinct, $(a-b)(b-c)(c-a)ne 0$. We conclude that
        $(abc)^2=1$. This yields the two possibilities $abc=1$ and $abc=-1$.



        In a logical sense we are finished: We have shown that if $(a,b,c)$ is a solution of the system with $a$, $b$, and $c$ distinct, then $abc=pm 1$.



        But it would be nice to show that there are solutions of the desired type. So let's find such a solution, with $abc=1$.



        Look for a solution with $a=1$. Then we need $c=1/b$. In order to satisfy our equations, we need $1+1/b=2b$. Beside the useless solution $b=1$, this has the solution $b=-1/2$. We conclude that
        $$a=1,quad b=-frac{1}{2},quad c=-2$$
        is a solution of the desired type, with $abc=1$. By changing all the signs, we find that
        $$a=-1,quad b=frac{1}{2},quad c=2$$
        is a solution with $abc=-1$.



        Added: It is easy to see that if two 0f $a$, $b$, $c$ are equal, then they are all equal, giving the parametric family $(t,t,t)$, where $tne 0$. Now assume that $abc=pm 1$. We find all solutions with $abc=1$. The solutions with $abc=-1$ are then obtained by changing all the signs.



        Let $a=t$. Our equations will be satisfied if $t+1/b=b+1/c$, where $c=1/bt$. We therefore obtain the equation $t+1/b=b+bt$, which simplifies to $(1+t)b^2 -tb-1=0$. Now we can solve this quadratic equation for $b$, and get $c$ from $tbc=1$. There is undoubtedly a more symmetric way to obtain the complete parametric description of the solutions!






        share|cite|improve this answer











        $endgroup$



        From the fact that Expressions $1$ and $2$ are equal, we obtain
        $$a-b=frac{1}{c}-frac{1}{b}=frac{b-c}{bc}.$$
        From the fact that Expressions $2$ and $3$ are equal, we obtain
        $$b-c=frac{1}{a}-frac{1}{c}=frac{c-a}{ca}.$$
        From the fact that Expressions $3$ and $1$ are equal, we obtain
        $$c-a=frac{1}{b}-frac{1}{a}=frac{a-b}{ab}.$$



        Multiply the left-hand sides, the right-hand sides. We get
        $$(a-b)(b-c)(c-a)=frac{(b-c)(c-a)(a-b)}{(abc)^2}.$$
        Since $a$, $b$, and $c$ are distinct, $(a-b)(b-c)(c-a)ne 0$. We conclude that
        $(abc)^2=1$. This yields the two possibilities $abc=1$ and $abc=-1$.



        In a logical sense we are finished: We have shown that if $(a,b,c)$ is a solution of the system with $a$, $b$, and $c$ distinct, then $abc=pm 1$.



        But it would be nice to show that there are solutions of the desired type. So let's find such a solution, with $abc=1$.



        Look for a solution with $a=1$. Then we need $c=1/b$. In order to satisfy our equations, we need $1+1/b=2b$. Beside the useless solution $b=1$, this has the solution $b=-1/2$. We conclude that
        $$a=1,quad b=-frac{1}{2},quad c=-2$$
        is a solution of the desired type, with $abc=1$. By changing all the signs, we find that
        $$a=-1,quad b=frac{1}{2},quad c=2$$
        is a solution with $abc=-1$.



        Added: It is easy to see that if two 0f $a$, $b$, $c$ are equal, then they are all equal, giving the parametric family $(t,t,t)$, where $tne 0$. Now assume that $abc=pm 1$. We find all solutions with $abc=1$. The solutions with $abc=-1$ are then obtained by changing all the signs.



        Let $a=t$. Our equations will be satisfied if $t+1/b=b+1/c$, where $c=1/bt$. We therefore obtain the equation $t+1/b=b+bt$, which simplifies to $(1+t)b^2 -tb-1=0$. Now we can solve this quadratic equation for $b$, and get $c$ from $tbc=1$. There is undoubtedly a more symmetric way to obtain the complete parametric description of the solutions!







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 1 '11 at 4:58

























        answered Sep 1 '11 at 0:10









        André NicolasAndré Nicolas

        453k36428813




        453k36428813












        • $begingroup$
          This method is exactly the same as that presented in Aryabhata's answer, except for the choice of notation: your $: a-b = (b-c)/(bc)$ vs. his $:(a-b):A = a:(b-c):$ for $A = abc:.:$ But the key point is to exploit the innate symmetry, which is more clearly presented above.
          $endgroup$
          – Bill Dubuque
          Sep 1 '11 at 2:20








        • 1




          $begingroup$
          @Bill Dubuque: After I had posted the solution, I noticed that Aryabhata must have posted a solution while I was typing. Wondered whether to delete, decided not to because the line of attack that I used, though essentially identical, is probably more natural from a student point of view. And yes, I did want to pound in the use of symmetry!
          $endgroup$
          – André Nicolas
          Sep 1 '11 at 2:56


















        • $begingroup$
          This method is exactly the same as that presented in Aryabhata's answer, except for the choice of notation: your $: a-b = (b-c)/(bc)$ vs. his $:(a-b):A = a:(b-c):$ for $A = abc:.:$ But the key point is to exploit the innate symmetry, which is more clearly presented above.
          $endgroup$
          – Bill Dubuque
          Sep 1 '11 at 2:20








        • 1




          $begingroup$
          @Bill Dubuque: After I had posted the solution, I noticed that Aryabhata must have posted a solution while I was typing. Wondered whether to delete, decided not to because the line of attack that I used, though essentially identical, is probably more natural from a student point of view. And yes, I did want to pound in the use of symmetry!
          $endgroup$
          – André Nicolas
          Sep 1 '11 at 2:56
















        $begingroup$
        This method is exactly the same as that presented in Aryabhata's answer, except for the choice of notation: your $: a-b = (b-c)/(bc)$ vs. his $:(a-b):A = a:(b-c):$ for $A = abc:.:$ But the key point is to exploit the innate symmetry, which is more clearly presented above.
        $endgroup$
        – Bill Dubuque
        Sep 1 '11 at 2:20






        $begingroup$
        This method is exactly the same as that presented in Aryabhata's answer, except for the choice of notation: your $: a-b = (b-c)/(bc)$ vs. his $:(a-b):A = a:(b-c):$ for $A = abc:.:$ But the key point is to exploit the innate symmetry, which is more clearly presented above.
        $endgroup$
        – Bill Dubuque
        Sep 1 '11 at 2:20






        1




        1




        $begingroup$
        @Bill Dubuque: After I had posted the solution, I noticed that Aryabhata must have posted a solution while I was typing. Wondered whether to delete, decided not to because the line of attack that I used, though essentially identical, is probably more natural from a student point of view. And yes, I did want to pound in the use of symmetry!
        $endgroup$
        – André Nicolas
        Sep 1 '11 at 2:56




        $begingroup$
        @Bill Dubuque: After I had posted the solution, I noticed that Aryabhata must have posted a solution while I was typing. Wondered whether to delete, decided not to because the line of attack that I used, though essentially identical, is probably more natural from a student point of view. And yes, I did want to pound in the use of symmetry!
        $endgroup$
        – André Nicolas
        Sep 1 '11 at 2:56











        14












        $begingroup$

        Give the common value of $a+1/b$ etc. a name, say $h$. We can now rewrite $a+1/b=h$ to $a=h-1/b$ and similarly $b=h-1/c$ and $c=h-1/a$. Telescoping these expressions into each other gives
        $$a=frac{h^2-ah-1}{h^3-ah^2-2h+a}$$
        which rearranges to
        $$(1-h^2)a^2 + (h^3-h)a + 1-h^2 = 0$$
        Now because everything is symmetric, $b$ and $c$ must satisfy the same equation, but $a$, $b$ and $c$ were distinct numbers, so they can only be roots in a quadratic polynomial if it is identically zero. So $h^2=1$, which makes all of the coefficients vanish. Thus $h=pm 1$.



        Assuming that $h=1$ we now get $b=frac{1}{1-a}$ and $c=1-frac{1}{a}=frac{a-1}{a}$, hence $abc=-1$. An example is $(a,b,c)=(2,-1,1/2)$.



        Negating everything in the $h=1$ case gives $h=-1$ and $abc=1$.






        share|cite|improve this answer











        $endgroup$


















          14












          $begingroup$

          Give the common value of $a+1/b$ etc. a name, say $h$. We can now rewrite $a+1/b=h$ to $a=h-1/b$ and similarly $b=h-1/c$ and $c=h-1/a$. Telescoping these expressions into each other gives
          $$a=frac{h^2-ah-1}{h^3-ah^2-2h+a}$$
          which rearranges to
          $$(1-h^2)a^2 + (h^3-h)a + 1-h^2 = 0$$
          Now because everything is symmetric, $b$ and $c$ must satisfy the same equation, but $a$, $b$ and $c$ were distinct numbers, so they can only be roots in a quadratic polynomial if it is identically zero. So $h^2=1$, which makes all of the coefficients vanish. Thus $h=pm 1$.



          Assuming that $h=1$ we now get $b=frac{1}{1-a}$ and $c=1-frac{1}{a}=frac{a-1}{a}$, hence $abc=-1$. An example is $(a,b,c)=(2,-1,1/2)$.



          Negating everything in the $h=1$ case gives $h=-1$ and $abc=1$.






          share|cite|improve this answer











          $endgroup$
















            14












            14








            14





            $begingroup$

            Give the common value of $a+1/b$ etc. a name, say $h$. We can now rewrite $a+1/b=h$ to $a=h-1/b$ and similarly $b=h-1/c$ and $c=h-1/a$. Telescoping these expressions into each other gives
            $$a=frac{h^2-ah-1}{h^3-ah^2-2h+a}$$
            which rearranges to
            $$(1-h^2)a^2 + (h^3-h)a + 1-h^2 = 0$$
            Now because everything is symmetric, $b$ and $c$ must satisfy the same equation, but $a$, $b$ and $c$ were distinct numbers, so they can only be roots in a quadratic polynomial if it is identically zero. So $h^2=1$, which makes all of the coefficients vanish. Thus $h=pm 1$.



            Assuming that $h=1$ we now get $b=frac{1}{1-a}$ and $c=1-frac{1}{a}=frac{a-1}{a}$, hence $abc=-1$. An example is $(a,b,c)=(2,-1,1/2)$.



            Negating everything in the $h=1$ case gives $h=-1$ and $abc=1$.






            share|cite|improve this answer











            $endgroup$



            Give the common value of $a+1/b$ etc. a name, say $h$. We can now rewrite $a+1/b=h$ to $a=h-1/b$ and similarly $b=h-1/c$ and $c=h-1/a$. Telescoping these expressions into each other gives
            $$a=frac{h^2-ah-1}{h^3-ah^2-2h+a}$$
            which rearranges to
            $$(1-h^2)a^2 + (h^3-h)a + 1-h^2 = 0$$
            Now because everything is symmetric, $b$ and $c$ must satisfy the same equation, but $a$, $b$ and $c$ were distinct numbers, so they can only be roots in a quadratic polynomial if it is identically zero. So $h^2=1$, which makes all of the coefficients vanish. Thus $h=pm 1$.



            Assuming that $h=1$ we now get $b=frac{1}{1-a}$ and $c=1-frac{1}{a}=frac{a-1}{a}$, hence $abc=-1$. An example is $(a,b,c)=(2,-1,1/2)$.



            Negating everything in the $h=1$ case gives $h=-1$ and $abc=1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 1 '11 at 1:28

























            answered Aug 31 '11 at 22:43









            Henning MakholmHenning Makholm

            240k17306544




            240k17306544























                12












                $begingroup$

                Let $A = abc$. We can assume $A ne 0$.



                Now the equations can be written as



                $a + ac/A = b + ab/A = c + bc/A$



                Multiply by $A$ throughout.



                $a(A+c) = b(A+a) = c(A+b)$



                Subtract first two, last two, first and last to get three equations and multiply those to get an equation in $A$.






                share|cite|improve this answer









                $endgroup$


















                  12












                  $begingroup$

                  Let $A = abc$. We can assume $A ne 0$.



                  Now the equations can be written as



                  $a + ac/A = b + ab/A = c + bc/A$



                  Multiply by $A$ throughout.



                  $a(A+c) = b(A+a) = c(A+b)$



                  Subtract first two, last two, first and last to get three equations and multiply those to get an equation in $A$.






                  share|cite|improve this answer









                  $endgroup$
















                    12












                    12








                    12





                    $begingroup$

                    Let $A = abc$. We can assume $A ne 0$.



                    Now the equations can be written as



                    $a + ac/A = b + ab/A = c + bc/A$



                    Multiply by $A$ throughout.



                    $a(A+c) = b(A+a) = c(A+b)$



                    Subtract first two, last two, first and last to get three equations and multiply those to get an equation in $A$.






                    share|cite|improve this answer









                    $endgroup$



                    Let $A = abc$. We can assume $A ne 0$.



                    Now the equations can be written as



                    $a + ac/A = b + ab/A = c + bc/A$



                    Multiply by $A$ throughout.



                    $a(A+c) = b(A+a) = c(A+b)$



                    Subtract first two, last two, first and last to get three equations and multiply those to get an equation in $A$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 1 '11 at 0:05









                    AryabhataAryabhata

                    70k6156246




                    70k6156246























                        7












                        $begingroup$

                        Below I explicitly highlight the symmetry implicit in the answers of Aryabhata and Andre. Let $rm:sigma:$ be the permutation $rm:(a b c):$ acting on polynomials in $rm:a,b,c:$ by cyclically permuting the variables, i.e. $rm:sigma f(a,b,c): =: f(b,c,a):, $ e.g. $rm sigma(a-b) = b-c:.:$ Define the norm by $rm:N(g) = g sigma g sigma^2 g:.:$ Note that $rm:N(sigma: g): =: N(g):$ since $rm:sigma^3 = 1:.:$ Put $rm:f: =: abc: =: N(a):,:$ so $rm:sigma: f: =: f:.:$ Put $rm:g: =: a-b:.:$ Subtracting each pair of the given three equations yields the following equations



                        $$begin{eqnarray*} rm f &=&rm a dfrac{b-c}{a-b} &=&rm a dfrac{sigma: g}{g}



                        rm f &=&rm b dfrac{c-a}{b-c} &=&rm :sigma a dfrac{sigma^2 g}{sigma: g}



                        rm f &=&rm c dfrac{a-b}{c-a} &=&rm sigma^2 a dfrac{sigma^3:g}{sigma^2g}
                        end{eqnarray*}$$



                        Multiplying the above, using $rm:sigma^3:g = g:,:$ yields $rm:f^{:3} = N(a) = f:,:$ so $rm:f ne 0:$ $:Rightarrow:$ $rm:f^{:2} = 1:.:$ That's essentially the method employed in said answers. I show that this method amounts to simply taking the norm of the first equation. Notice that the second and third equations result from successively applying $rm:sigma:$ to the first equation, i.e. the given three equations are equivalent to stating that the first equation is preserved by $rm:sigma:$ and $rm:sigma^2:$. So multiplying them all amounts to taking the norm of the first, i.e.



                        $$rm f^{:3} = N(f) = Nbigg(a:dfrac{sigma g}{g}bigg) = N(a) = f quad by quad N(sigma g) = N(g) $$



                        So, from this viewpoint, the proof that $rm:f^{:3} = f:$ is a one-line inference achieved by by taking a norm (that $rm:f/a = sigma:g/g:$ has norm $1$ is essentially multiplicative telescopy). This simplicity above results from recognizing and exploiting the innate symmetry in the problem.



                        Less trivial exploitation of similar symmetries arise in the Galois theory of difference equations and radical extensions (Kummer theory). For one simple example see this answer.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          Nice answer, bringing out the structure explicitly.
                          $endgroup$
                          – André Nicolas
                          Sep 1 '11 at 4:53
















                        7












                        $begingroup$

                        Below I explicitly highlight the symmetry implicit in the answers of Aryabhata and Andre. Let $rm:sigma:$ be the permutation $rm:(a b c):$ acting on polynomials in $rm:a,b,c:$ by cyclically permuting the variables, i.e. $rm:sigma f(a,b,c): =: f(b,c,a):, $ e.g. $rm sigma(a-b) = b-c:.:$ Define the norm by $rm:N(g) = g sigma g sigma^2 g:.:$ Note that $rm:N(sigma: g): =: N(g):$ since $rm:sigma^3 = 1:.:$ Put $rm:f: =: abc: =: N(a):,:$ so $rm:sigma: f: =: f:.:$ Put $rm:g: =: a-b:.:$ Subtracting each pair of the given three equations yields the following equations



                        $$begin{eqnarray*} rm f &=&rm a dfrac{b-c}{a-b} &=&rm a dfrac{sigma: g}{g}



                        rm f &=&rm b dfrac{c-a}{b-c} &=&rm :sigma a dfrac{sigma^2 g}{sigma: g}



                        rm f &=&rm c dfrac{a-b}{c-a} &=&rm sigma^2 a dfrac{sigma^3:g}{sigma^2g}
                        end{eqnarray*}$$



                        Multiplying the above, using $rm:sigma^3:g = g:,:$ yields $rm:f^{:3} = N(a) = f:,:$ so $rm:f ne 0:$ $:Rightarrow:$ $rm:f^{:2} = 1:.:$ That's essentially the method employed in said answers. I show that this method amounts to simply taking the norm of the first equation. Notice that the second and third equations result from successively applying $rm:sigma:$ to the first equation, i.e. the given three equations are equivalent to stating that the first equation is preserved by $rm:sigma:$ and $rm:sigma^2:$. So multiplying them all amounts to taking the norm of the first, i.e.



                        $$rm f^{:3} = N(f) = Nbigg(a:dfrac{sigma g}{g}bigg) = N(a) = f quad by quad N(sigma g) = N(g) $$



                        So, from this viewpoint, the proof that $rm:f^{:3} = f:$ is a one-line inference achieved by by taking a norm (that $rm:f/a = sigma:g/g:$ has norm $1$ is essentially multiplicative telescopy). This simplicity above results from recognizing and exploiting the innate symmetry in the problem.



                        Less trivial exploitation of similar symmetries arise in the Galois theory of difference equations and radical extensions (Kummer theory). For one simple example see this answer.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          Nice answer, bringing out the structure explicitly.
                          $endgroup$
                          – André Nicolas
                          Sep 1 '11 at 4:53














                        7












                        7








                        7





                        $begingroup$

                        Below I explicitly highlight the symmetry implicit in the answers of Aryabhata and Andre. Let $rm:sigma:$ be the permutation $rm:(a b c):$ acting on polynomials in $rm:a,b,c:$ by cyclically permuting the variables, i.e. $rm:sigma f(a,b,c): =: f(b,c,a):, $ e.g. $rm sigma(a-b) = b-c:.:$ Define the norm by $rm:N(g) = g sigma g sigma^2 g:.:$ Note that $rm:N(sigma: g): =: N(g):$ since $rm:sigma^3 = 1:.:$ Put $rm:f: =: abc: =: N(a):,:$ so $rm:sigma: f: =: f:.:$ Put $rm:g: =: a-b:.:$ Subtracting each pair of the given three equations yields the following equations



                        $$begin{eqnarray*} rm f &=&rm a dfrac{b-c}{a-b} &=&rm a dfrac{sigma: g}{g}



                        rm f &=&rm b dfrac{c-a}{b-c} &=&rm :sigma a dfrac{sigma^2 g}{sigma: g}



                        rm f &=&rm c dfrac{a-b}{c-a} &=&rm sigma^2 a dfrac{sigma^3:g}{sigma^2g}
                        end{eqnarray*}$$



                        Multiplying the above, using $rm:sigma^3:g = g:,:$ yields $rm:f^{:3} = N(a) = f:,:$ so $rm:f ne 0:$ $:Rightarrow:$ $rm:f^{:2} = 1:.:$ That's essentially the method employed in said answers. I show that this method amounts to simply taking the norm of the first equation. Notice that the second and third equations result from successively applying $rm:sigma:$ to the first equation, i.e. the given three equations are equivalent to stating that the first equation is preserved by $rm:sigma:$ and $rm:sigma^2:$. So multiplying them all amounts to taking the norm of the first, i.e.



                        $$rm f^{:3} = N(f) = Nbigg(a:dfrac{sigma g}{g}bigg) = N(a) = f quad by quad N(sigma g) = N(g) $$



                        So, from this viewpoint, the proof that $rm:f^{:3} = f:$ is a one-line inference achieved by by taking a norm (that $rm:f/a = sigma:g/g:$ has norm $1$ is essentially multiplicative telescopy). This simplicity above results from recognizing and exploiting the innate symmetry in the problem.



                        Less trivial exploitation of similar symmetries arise in the Galois theory of difference equations and radical extensions (Kummer theory). For one simple example see this answer.






                        share|cite|improve this answer











                        $endgroup$



                        Below I explicitly highlight the symmetry implicit in the answers of Aryabhata and Andre. Let $rm:sigma:$ be the permutation $rm:(a b c):$ acting on polynomials in $rm:a,b,c:$ by cyclically permuting the variables, i.e. $rm:sigma f(a,b,c): =: f(b,c,a):, $ e.g. $rm sigma(a-b) = b-c:.:$ Define the norm by $rm:N(g) = g sigma g sigma^2 g:.:$ Note that $rm:N(sigma: g): =: N(g):$ since $rm:sigma^3 = 1:.:$ Put $rm:f: =: abc: =: N(a):,:$ so $rm:sigma: f: =: f:.:$ Put $rm:g: =: a-b:.:$ Subtracting each pair of the given three equations yields the following equations



                        $$begin{eqnarray*} rm f &=&rm a dfrac{b-c}{a-b} &=&rm a dfrac{sigma: g}{g}



                        rm f &=&rm b dfrac{c-a}{b-c} &=&rm :sigma a dfrac{sigma^2 g}{sigma: g}



                        rm f &=&rm c dfrac{a-b}{c-a} &=&rm sigma^2 a dfrac{sigma^3:g}{sigma^2g}
                        end{eqnarray*}$$



                        Multiplying the above, using $rm:sigma^3:g = g:,:$ yields $rm:f^{:3} = N(a) = f:,:$ so $rm:f ne 0:$ $:Rightarrow:$ $rm:f^{:2} = 1:.:$ That's essentially the method employed in said answers. I show that this method amounts to simply taking the norm of the first equation. Notice that the second and third equations result from successively applying $rm:sigma:$ to the first equation, i.e. the given three equations are equivalent to stating that the first equation is preserved by $rm:sigma:$ and $rm:sigma^2:$. So multiplying them all amounts to taking the norm of the first, i.e.



                        $$rm f^{:3} = N(f) = Nbigg(a:dfrac{sigma g}{g}bigg) = N(a) = f quad by quad N(sigma g) = N(g) $$



                        So, from this viewpoint, the proof that $rm:f^{:3} = f:$ is a one-line inference achieved by by taking a norm (that $rm:f/a = sigma:g/g:$ has norm $1$ is essentially multiplicative telescopy). This simplicity above results from recognizing and exploiting the innate symmetry in the problem.



                        Less trivial exploitation of similar symmetries arise in the Galois theory of difference equations and radical extensions (Kummer theory). For one simple example see this answer.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Apr 13 '17 at 12:20









                        Community

                        1




                        1










                        answered Sep 1 '11 at 4:43









                        Bill DubuqueBill Dubuque

                        211k29193646




                        211k29193646












                        • $begingroup$
                          Nice answer, bringing out the structure explicitly.
                          $endgroup$
                          – André Nicolas
                          Sep 1 '11 at 4:53


















                        • $begingroup$
                          Nice answer, bringing out the structure explicitly.
                          $endgroup$
                          – André Nicolas
                          Sep 1 '11 at 4:53
















                        $begingroup$
                        Nice answer, bringing out the structure explicitly.
                        $endgroup$
                        – André Nicolas
                        Sep 1 '11 at 4:53




                        $begingroup$
                        Nice answer, bringing out the structure explicitly.
                        $endgroup$
                        – André Nicolas
                        Sep 1 '11 at 4:53











                        0












                        $begingroup$

                        I'm not sure how different this is, but here is my version



                        $a + frac{1}{b} = b + frac{1}{c} = c + frac{1}{a}
                        quad$ (Note this implies $abc ne 0$)



                        $a^2bc + ac = ab^2c + ab = abc^2 + bc$



                        $a(abc) + ac = b(abc) + ab = c(abc) + bc$



                        $quad a(abc) + ac = b(abc) + ab implies (a-b)abc=a(b-c)$

                        $$ a=b=c ne 0 ; text{ or } ; abc = dfrac{a(b-c)}{a-b}$$



                        $quad a(abc) + ac = c(abc) + bc implies (a-c)abc = c(b-a)$
                        $$a=c=b ne 0 ; text{ or } ; abc = dfrac{c(b-a)}{a-c}$$



                        $quad b(abc) + ab = c(abc) + bc implies (b-c)abc=b(c-a)$
                        $$b=c=a ne 0 ; text{ or } ; abc = dfrac{b(c-a)}{b-c}$$



                        So, one solution is $; a=b=c ne 0$.

                        But if $a,b,$ and $c$ are distinct and non zero, then



                        $(abc)^3 =
                        dfrac{a(b-c)}{a-b} cdot dfrac{c(b-a)}{a-c} cdot dfrac{b(c-a)}{b-c}$



                        $(abc)^3 = abc$



                        $(abc)^2 = 1$



                        $abc = pm 1$



                        $|abc| = 1$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          I'm not sure how different this is, but here is my version



                          $a + frac{1}{b} = b + frac{1}{c} = c + frac{1}{a}
                          quad$ (Note this implies $abc ne 0$)



                          $a^2bc + ac = ab^2c + ab = abc^2 + bc$



                          $a(abc) + ac = b(abc) + ab = c(abc) + bc$



                          $quad a(abc) + ac = b(abc) + ab implies (a-b)abc=a(b-c)$

                          $$ a=b=c ne 0 ; text{ or } ; abc = dfrac{a(b-c)}{a-b}$$



                          $quad a(abc) + ac = c(abc) + bc implies (a-c)abc = c(b-a)$
                          $$a=c=b ne 0 ; text{ or } ; abc = dfrac{c(b-a)}{a-c}$$



                          $quad b(abc) + ab = c(abc) + bc implies (b-c)abc=b(c-a)$
                          $$b=c=a ne 0 ; text{ or } ; abc = dfrac{b(c-a)}{b-c}$$



                          So, one solution is $; a=b=c ne 0$.

                          But if $a,b,$ and $c$ are distinct and non zero, then



                          $(abc)^3 =
                          dfrac{a(b-c)}{a-b} cdot dfrac{c(b-a)}{a-c} cdot dfrac{b(c-a)}{b-c}$



                          $(abc)^3 = abc$



                          $(abc)^2 = 1$



                          $abc = pm 1$



                          $|abc| = 1$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            I'm not sure how different this is, but here is my version



                            $a + frac{1}{b} = b + frac{1}{c} = c + frac{1}{a}
                            quad$ (Note this implies $abc ne 0$)



                            $a^2bc + ac = ab^2c + ab = abc^2 + bc$



                            $a(abc) + ac = b(abc) + ab = c(abc) + bc$



                            $quad a(abc) + ac = b(abc) + ab implies (a-b)abc=a(b-c)$

                            $$ a=b=c ne 0 ; text{ or } ; abc = dfrac{a(b-c)}{a-b}$$



                            $quad a(abc) + ac = c(abc) + bc implies (a-c)abc = c(b-a)$
                            $$a=c=b ne 0 ; text{ or } ; abc = dfrac{c(b-a)}{a-c}$$



                            $quad b(abc) + ab = c(abc) + bc implies (b-c)abc=b(c-a)$
                            $$b=c=a ne 0 ; text{ or } ; abc = dfrac{b(c-a)}{b-c}$$



                            So, one solution is $; a=b=c ne 0$.

                            But if $a,b,$ and $c$ are distinct and non zero, then



                            $(abc)^3 =
                            dfrac{a(b-c)}{a-b} cdot dfrac{c(b-a)}{a-c} cdot dfrac{b(c-a)}{b-c}$



                            $(abc)^3 = abc$



                            $(abc)^2 = 1$



                            $abc = pm 1$



                            $|abc| = 1$






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                            $endgroup$



                            I'm not sure how different this is, but here is my version



                            $a + frac{1}{b} = b + frac{1}{c} = c + frac{1}{a}
                            quad$ (Note this implies $abc ne 0$)



                            $a^2bc + ac = ab^2c + ab = abc^2 + bc$



                            $a(abc) + ac = b(abc) + ab = c(abc) + bc$



                            $quad a(abc) + ac = b(abc) + ab implies (a-b)abc=a(b-c)$

                            $$ a=b=c ne 0 ; text{ or } ; abc = dfrac{a(b-c)}{a-b}$$



                            $quad a(abc) + ac = c(abc) + bc implies (a-c)abc = c(b-a)$
                            $$a=c=b ne 0 ; text{ or } ; abc = dfrac{c(b-a)}{a-c}$$



                            $quad b(abc) + ab = c(abc) + bc implies (b-c)abc=b(c-a)$
                            $$b=c=a ne 0 ; text{ or } ; abc = dfrac{b(c-a)}{b-c}$$



                            So, one solution is $; a=b=c ne 0$.

                            But if $a,b,$ and $c$ are distinct and non zero, then



                            $(abc)^3 =
                            dfrac{a(b-c)}{a-b} cdot dfrac{c(b-a)}{a-c} cdot dfrac{b(c-a)}{b-c}$



                            $(abc)^3 = abc$



                            $(abc)^2 = 1$



                            $abc = pm 1$



                            $|abc| = 1$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Sep 4 '15 at 14:55









                            steven gregorysteven gregory

                            18.2k32258




                            18.2k32258






























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