How do I find all prime solutions $p, q, r$ of the equation $displaystyle p(p+1)+q(q+1) = r(r+1)$?











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Find primes $p, q, r$ of the equation $$p(p+1)+q(q+1)
= r(r+1)$$




I know that it has only one solution namely $p = q = 2,r = 3$. But i can't show that.



Thank you for any help










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  • How do you know that no other solutions exist then, what is your source ?
    – Peter
    Dec 12 '16 at 22:28










  • I think this problem related to A shinzel solution of the titled equation if i'm true
    – zeraoulia rafik
    Dec 12 '16 at 22:43










  • It may or may not help to rephrase it as $p^2 + p + q^2 + q = r^2 + r$.
    – Robert Soupe
    Dec 13 '16 at 2:21












  • this equivalent to solve $$(2p+1)^2+(2q+1)^2=(2r+1)^2+1$$
    – Bumblebee
    Jan 18 at 20:01















up vote
5
down vote

favorite
2













Find primes $p, q, r$ of the equation $$p(p+1)+q(q+1)
= r(r+1)$$




I know that it has only one solution namely $p = q = 2,r = 3$. But i can't show that.



Thank you for any help










share|cite|improve this question
























  • How do you know that no other solutions exist then, what is your source ?
    – Peter
    Dec 12 '16 at 22:28










  • I think this problem related to A shinzel solution of the titled equation if i'm true
    – zeraoulia rafik
    Dec 12 '16 at 22:43










  • It may or may not help to rephrase it as $p^2 + p + q^2 + q = r^2 + r$.
    – Robert Soupe
    Dec 13 '16 at 2:21












  • this equivalent to solve $$(2p+1)^2+(2q+1)^2=(2r+1)^2+1$$
    – Bumblebee
    Jan 18 at 20:01













up vote
5
down vote

favorite
2









up vote
5
down vote

favorite
2






2






Find primes $p, q, r$ of the equation $$p(p+1)+q(q+1)
= r(r+1)$$




I know that it has only one solution namely $p = q = 2,r = 3$. But i can't show that.



Thank you for any help










share|cite|improve this question
















Find primes $p, q, r$ of the equation $$p(p+1)+q(q+1)
= r(r+1)$$




I know that it has only one solution namely $p = q = 2,r = 3$. But i can't show that.



Thank you for any help







number-theory elementary-number-theory prime-numbers






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edited Nov 15 at 17:25









greedoid

34.4k114488




34.4k114488










asked Dec 12 '16 at 22:12









Youssra El Yossra Youssra

975




975












  • How do you know that no other solutions exist then, what is your source ?
    – Peter
    Dec 12 '16 at 22:28










  • I think this problem related to A shinzel solution of the titled equation if i'm true
    – zeraoulia rafik
    Dec 12 '16 at 22:43










  • It may or may not help to rephrase it as $p^2 + p + q^2 + q = r^2 + r$.
    – Robert Soupe
    Dec 13 '16 at 2:21












  • this equivalent to solve $$(2p+1)^2+(2q+1)^2=(2r+1)^2+1$$
    – Bumblebee
    Jan 18 at 20:01


















  • How do you know that no other solutions exist then, what is your source ?
    – Peter
    Dec 12 '16 at 22:28










  • I think this problem related to A shinzel solution of the titled equation if i'm true
    – zeraoulia rafik
    Dec 12 '16 at 22:43










  • It may or may not help to rephrase it as $p^2 + p + q^2 + q = r^2 + r$.
    – Robert Soupe
    Dec 13 '16 at 2:21












  • this equivalent to solve $$(2p+1)^2+(2q+1)^2=(2r+1)^2+1$$
    – Bumblebee
    Jan 18 at 20:01
















How do you know that no other solutions exist then, what is your source ?
– Peter
Dec 12 '16 at 22:28




How do you know that no other solutions exist then, what is your source ?
– Peter
Dec 12 '16 at 22:28












I think this problem related to A shinzel solution of the titled equation if i'm true
– zeraoulia rafik
Dec 12 '16 at 22:43




I think this problem related to A shinzel solution of the titled equation if i'm true
– zeraoulia rafik
Dec 12 '16 at 22:43












It may or may not help to rephrase it as $p^2 + p + q^2 + q = r^2 + r$.
– Robert Soupe
Dec 13 '16 at 2:21






It may or may not help to rephrase it as $p^2 + p + q^2 + q = r^2 + r$.
– Robert Soupe
Dec 13 '16 at 2:21














this equivalent to solve $$(2p+1)^2+(2q+1)^2=(2r+1)^2+1$$
– Bumblebee
Jan 18 at 20:01




this equivalent to solve $$(2p+1)^2+(2q+1)^2=(2r+1)^2+1$$
– Bumblebee
Jan 18 at 20:01










3 Answers
3






active

oldest

votes

















up vote
7
down vote



accepted










May this lead to a simple proof for your problem according to your unic example of solution . There is only one
solution, namely $p = q = 2,r = 3$. To see that, we shall find all solutions
of the equation $p(p+1)+q(q+1) = n(n+1)$ where $p$ and $q$ are primes and
$n$ is a positive integer. Our equation yields
$p(p+1) = n(n+1)-q(q+1) = (n-q)(n+q+1)$,
and we must have $n > q$. Since $p$ is a prime, we have either $p|n-q$ or
$p|n+q+1$. If $p|n-q$, then we have $pleq n-q$, which implies $p(p+1)
leq (n-q)(n-q+1)$, and therefore $n+q+1 leq n-q+1$, which is impossible.
Thus we have $p|n+q+ 1$, which means that for some positive integer $k$
,$n+q+1 = kp$, which implies $p+1 = k(n-q)tag1$.



If we had $k = 1$, then $n+q+ 1 = p$ and $p+ 1 = n-q$, which gives $p-q
= n+ 1$ and $p+q = n- 1$, which is impossible, because $(p+q)-(p-q)=2q>0$ and $(n-1)-(n+1)=-2<0$. Thus, $k > 1$. From $(1)$ we
easily obtain:



$2q = (n+q)-(n-q) = kp-1-(n-q)
= k[k(n-q)-1]-1-(n-q) = (k+1)[(k-1)(n-q)-1]$



.Since $k geq 2$, we have $k+1 geq 3$. The last equality, whose left-hand side has
positive integer divisors $1, 2, q$, and $2q$ only, implies that either $k+ 1 = q$
or $k+1 = 2q$. If $k+1 = q$, then $(k-1)(n-q) = 3$, hence $(q-2)(n-q) = 3$.
This leads to either $q-2 = 1$, $n-q = 3$, that is $q = 3, n = 6, k = q-1 = 2$,
and, in view of $(1)$, $P = 5$, or else, $q-2 = 3$, $n-q = 1$, which gives $q = 5,
n = 6, k = 4$, and in view of $(1)$, $p = 3$.
On the other hand, if $k+1 = 2q$, then $(k-1)(n-q) = 2$, hence
$2(q-1)(n-q) = 2$. This leads to $q-1 = 1$ and $n-q = 1$, or $q = 2, n = 3$,
and, in view of $(1)$, $p = 2$. Thus, for positive integer $ n$, we have the
following solutions in primes $p$ and $q$: $(p = q = 2, n = 3; 2)$,$ (p = 5,
q = 3, n = 6)$, and $3$ ,$(p = 3, q = 5, n = 6)$. Only in the first solution all
three numbers are primes.



Note: If we denote by $displaystyle t_n = frac{n(n+1)}{2}$ the nth triangular number,
then the equation $t_p+t_q = t_r$
has only one solution in prime numbers, namely $p = q = 2, r = 3$.






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  • It should be $p+q=n-1$.
    – user236182
    Dec 12 '16 at 22:51










  • It's still impossible because $(p+q)-(p-q)=2q>0$ while $(n-1)-(n+1)=-2<0$.
    – user236182
    Dec 12 '16 at 22:54










  • for k=1 it's impossible and i cited your case p+q=n-1 in the side of n+q=p-1
    – zeraoulia rafik
    Dec 12 '16 at 22:55










  • If $p+1=n-q$, then $p+q=n-1$.
    – user236182
    Dec 12 '16 at 22:57










  • And if $n+q+1=p$, then $p-q=n+1$.
    – user236182
    Dec 12 '16 at 22:58


















up vote
0
down vote













Case $pne q$: We can assume that $p>q$, then $$r^2+r leq (p-1)^2+p^2+p-1+p=2p^2implies boxed{rleq psqrt{2}}$$



From:



$$p(p+1) =r^2- q^2+r-q =(r-q)(r+q+1)$$
we get



If $pmid r-q$ then $r+q+1mid p+1$, so $pleq r-q<r$ and $r+q+1leq p+1 implies r<p$ a contradiction.



If $pmid r+q+1$ then $r-qmid p+1$. Since $r+q+1=kp$ for some integer $kgeq 1$ we have $$kpleq psqrt{2}+(p-1)+1 implies kleq 2$$



Case 1: $k=2$ we get $r+q+1=2p$ and $p+1=2r-2q$ from where we get $3p=11$, no good.



Case 2: $k=1$ we get $r+q+1=p$ and $p+1=r-q$ so $q=-1$ again contradiction.





So $p=q$ and now we have to solve $$2p^2+2p = r^2+r$$



So $$2p(p+1)=r(r+1)implies rmid 2p(p+1)$$



If $rmid 2$ then $r=2$ which is impossibile.



If $rmid p$ then $rleq p$ which is impossibile.



If $rmid p+1$ then $rleq p+1$ but then $r=p+1$ since $r>p$, so $r=3$ and $p=2$.






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    up vote
    0
    down vote













    From the given equation
    $$p(p+1)+q(q+1)=r(r+1)$$
    it follows that $p < r$ and $q < r$.



    Next, another inequality which will be useful later . . .



    Claim:$;p+q > r$.



    Proof:
    begin{align*}
    &p(p+1)+q(q+1)=r(r+1)\[4pt]
    implies;&(p^2+q^2)+(p+q)=r(r+1)\[4pt]
    implies;&(p+q)^2+(p+q)>r(r+1)\[4pt]
    implies;&(p+q)(p+q+1)>r(r+1)\[4pt]
    implies;&p+q>r\[4pt]
    end{align*}

    as claimed.



    Returning to the main problem . . .



    First suppose $p=q$.



    Then the given equation reduces to
    $$2p(p+1)=r(r+1)$$
    hence, since $r > p$, it follows that $r|(p+1)$.



    But then $p < r le p+1$, so $r=p+1$, which implies $p=2$, and $r=3$.



    It can be verified that the triple $(p,q,r)=(2,2,3)$ satisfies the given equation.



    Next suppose $p,q$ are distinct.



    Without loss of generality, assume $p < q$.



    Suppose $;p=2$.



    Then from $p < q < r$, we get $qge 3$ and $rge 5$, hence
    begin{align*}
    &p(p+1)+q(q+1)=r(r+1)\[4pt]
    implies;&(2)(2+1)+q(q+1)=r(r+1)\[4pt]
    implies;&r(r+1)-q(q+1)=6\[4pt]
    implies;&(r-q)(q+r+1)=6\[4pt]
    implies;&(q+r+1)mid 6\[4pt]
    implies;&q+r+1le 6\[4pt]
    end{align*}

    contradiction, since $q+r+1ge 3+5+1=9$.



    Hence we must have $p > 2$.



    Since $p+q > r$, it follows that $pnotmid r-q$, and $qnotmid r-p$.
    begin{align*}
    text{Then};;&p(p+1)+q(q+1)=r(r+1)\[4pt]
    implies;&p(p+1)=r(r+1)-q(q+1)\[4pt]
    implies;&p(p+1)=(q+r+1)(r-q)\[4pt]
    implies;&pmid (q+r+1)\[4pt]
    implies;&pmid (p+q+r+1)\[12pt]
    text{and};,&p(p+1)+q(q+1)=r(r+1)\[4pt]
    implies;&q(q+1)=r(r+1)-p(p+1)\[4pt]
    implies;&q(q+1)=(p+r+1)(r-p)\[4pt]
    implies;&qmid (p+r+1)\[4pt]
    implies;&qmid (p+q+r+1)\[12pt]
    text{hence};,&pqmid (p+q+r+1)\[4pt]
    implies;&pqle p+q+r+1\[4pt]
    implies;&pq < p+q+(p+q)+1\[4pt]
    implies;&pq-2p-2q < 1\[4pt]
    implies;&(p-2)(q-2) < 5\[4pt]
    implies;&q-2 < 5\[4pt]
    implies;&q < 7\[4pt]
    implies;&qle 5\[4pt]
    implies;&(p,q)=(3,5)\[4pt]
    implies;&(3)(3+1)+(5)(5+1)=r(r+1)\[4pt]
    implies;&r=6\[4pt]
    end{align*}

    contradiction, since $6$ is not prime.



    Therefore the only solution is $(p,q,r)=(2,2,3)$.






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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      7
      down vote



      accepted










      May this lead to a simple proof for your problem according to your unic example of solution . There is only one
      solution, namely $p = q = 2,r = 3$. To see that, we shall find all solutions
      of the equation $p(p+1)+q(q+1) = n(n+1)$ where $p$ and $q$ are primes and
      $n$ is a positive integer. Our equation yields
      $p(p+1) = n(n+1)-q(q+1) = (n-q)(n+q+1)$,
      and we must have $n > q$. Since $p$ is a prime, we have either $p|n-q$ or
      $p|n+q+1$. If $p|n-q$, then we have $pleq n-q$, which implies $p(p+1)
      leq (n-q)(n-q+1)$, and therefore $n+q+1 leq n-q+1$, which is impossible.
      Thus we have $p|n+q+ 1$, which means that for some positive integer $k$
      ,$n+q+1 = kp$, which implies $p+1 = k(n-q)tag1$.



      If we had $k = 1$, then $n+q+ 1 = p$ and $p+ 1 = n-q$, which gives $p-q
      = n+ 1$ and $p+q = n- 1$, which is impossible, because $(p+q)-(p-q)=2q>0$ and $(n-1)-(n+1)=-2<0$. Thus, $k > 1$. From $(1)$ we
      easily obtain:



      $2q = (n+q)-(n-q) = kp-1-(n-q)
      = k[k(n-q)-1]-1-(n-q) = (k+1)[(k-1)(n-q)-1]$



      .Since $k geq 2$, we have $k+1 geq 3$. The last equality, whose left-hand side has
      positive integer divisors $1, 2, q$, and $2q$ only, implies that either $k+ 1 = q$
      or $k+1 = 2q$. If $k+1 = q$, then $(k-1)(n-q) = 3$, hence $(q-2)(n-q) = 3$.
      This leads to either $q-2 = 1$, $n-q = 3$, that is $q = 3, n = 6, k = q-1 = 2$,
      and, in view of $(1)$, $P = 5$, or else, $q-2 = 3$, $n-q = 1$, which gives $q = 5,
      n = 6, k = 4$, and in view of $(1)$, $p = 3$.
      On the other hand, if $k+1 = 2q$, then $(k-1)(n-q) = 2$, hence
      $2(q-1)(n-q) = 2$. This leads to $q-1 = 1$ and $n-q = 1$, or $q = 2, n = 3$,
      and, in view of $(1)$, $p = 2$. Thus, for positive integer $ n$, we have the
      following solutions in primes $p$ and $q$: $(p = q = 2, n = 3; 2)$,$ (p = 5,
      q = 3, n = 6)$, and $3$ ,$(p = 3, q = 5, n = 6)$. Only in the first solution all
      three numbers are primes.



      Note: If we denote by $displaystyle t_n = frac{n(n+1)}{2}$ the nth triangular number,
      then the equation $t_p+t_q = t_r$
      has only one solution in prime numbers, namely $p = q = 2, r = 3$.






      share|cite|improve this answer























      • It should be $p+q=n-1$.
        – user236182
        Dec 12 '16 at 22:51










      • It's still impossible because $(p+q)-(p-q)=2q>0$ while $(n-1)-(n+1)=-2<0$.
        – user236182
        Dec 12 '16 at 22:54










      • for k=1 it's impossible and i cited your case p+q=n-1 in the side of n+q=p-1
        – zeraoulia rafik
        Dec 12 '16 at 22:55










      • If $p+1=n-q$, then $p+q=n-1$.
        – user236182
        Dec 12 '16 at 22:57










      • And if $n+q+1=p$, then $p-q=n+1$.
        – user236182
        Dec 12 '16 at 22:58















      up vote
      7
      down vote



      accepted










      May this lead to a simple proof for your problem according to your unic example of solution . There is only one
      solution, namely $p = q = 2,r = 3$. To see that, we shall find all solutions
      of the equation $p(p+1)+q(q+1) = n(n+1)$ where $p$ and $q$ are primes and
      $n$ is a positive integer. Our equation yields
      $p(p+1) = n(n+1)-q(q+1) = (n-q)(n+q+1)$,
      and we must have $n > q$. Since $p$ is a prime, we have either $p|n-q$ or
      $p|n+q+1$. If $p|n-q$, then we have $pleq n-q$, which implies $p(p+1)
      leq (n-q)(n-q+1)$, and therefore $n+q+1 leq n-q+1$, which is impossible.
      Thus we have $p|n+q+ 1$, which means that for some positive integer $k$
      ,$n+q+1 = kp$, which implies $p+1 = k(n-q)tag1$.



      If we had $k = 1$, then $n+q+ 1 = p$ and $p+ 1 = n-q$, which gives $p-q
      = n+ 1$ and $p+q = n- 1$, which is impossible, because $(p+q)-(p-q)=2q>0$ and $(n-1)-(n+1)=-2<0$. Thus, $k > 1$. From $(1)$ we
      easily obtain:



      $2q = (n+q)-(n-q) = kp-1-(n-q)
      = k[k(n-q)-1]-1-(n-q) = (k+1)[(k-1)(n-q)-1]$



      .Since $k geq 2$, we have $k+1 geq 3$. The last equality, whose left-hand side has
      positive integer divisors $1, 2, q$, and $2q$ only, implies that either $k+ 1 = q$
      or $k+1 = 2q$. If $k+1 = q$, then $(k-1)(n-q) = 3$, hence $(q-2)(n-q) = 3$.
      This leads to either $q-2 = 1$, $n-q = 3$, that is $q = 3, n = 6, k = q-1 = 2$,
      and, in view of $(1)$, $P = 5$, or else, $q-2 = 3$, $n-q = 1$, which gives $q = 5,
      n = 6, k = 4$, and in view of $(1)$, $p = 3$.
      On the other hand, if $k+1 = 2q$, then $(k-1)(n-q) = 2$, hence
      $2(q-1)(n-q) = 2$. This leads to $q-1 = 1$ and $n-q = 1$, or $q = 2, n = 3$,
      and, in view of $(1)$, $p = 2$. Thus, for positive integer $ n$, we have the
      following solutions in primes $p$ and $q$: $(p = q = 2, n = 3; 2)$,$ (p = 5,
      q = 3, n = 6)$, and $3$ ,$(p = 3, q = 5, n = 6)$. Only in the first solution all
      three numbers are primes.



      Note: If we denote by $displaystyle t_n = frac{n(n+1)}{2}$ the nth triangular number,
      then the equation $t_p+t_q = t_r$
      has only one solution in prime numbers, namely $p = q = 2, r = 3$.






      share|cite|improve this answer























      • It should be $p+q=n-1$.
        – user236182
        Dec 12 '16 at 22:51










      • It's still impossible because $(p+q)-(p-q)=2q>0$ while $(n-1)-(n+1)=-2<0$.
        – user236182
        Dec 12 '16 at 22:54










      • for k=1 it's impossible and i cited your case p+q=n-1 in the side of n+q=p-1
        – zeraoulia rafik
        Dec 12 '16 at 22:55










      • If $p+1=n-q$, then $p+q=n-1$.
        – user236182
        Dec 12 '16 at 22:57










      • And if $n+q+1=p$, then $p-q=n+1$.
        – user236182
        Dec 12 '16 at 22:58













      up vote
      7
      down vote



      accepted







      up vote
      7
      down vote



      accepted






      May this lead to a simple proof for your problem according to your unic example of solution . There is only one
      solution, namely $p = q = 2,r = 3$. To see that, we shall find all solutions
      of the equation $p(p+1)+q(q+1) = n(n+1)$ where $p$ and $q$ are primes and
      $n$ is a positive integer. Our equation yields
      $p(p+1) = n(n+1)-q(q+1) = (n-q)(n+q+1)$,
      and we must have $n > q$. Since $p$ is a prime, we have either $p|n-q$ or
      $p|n+q+1$. If $p|n-q$, then we have $pleq n-q$, which implies $p(p+1)
      leq (n-q)(n-q+1)$, and therefore $n+q+1 leq n-q+1$, which is impossible.
      Thus we have $p|n+q+ 1$, which means that for some positive integer $k$
      ,$n+q+1 = kp$, which implies $p+1 = k(n-q)tag1$.



      If we had $k = 1$, then $n+q+ 1 = p$ and $p+ 1 = n-q$, which gives $p-q
      = n+ 1$ and $p+q = n- 1$, which is impossible, because $(p+q)-(p-q)=2q>0$ and $(n-1)-(n+1)=-2<0$. Thus, $k > 1$. From $(1)$ we
      easily obtain:



      $2q = (n+q)-(n-q) = kp-1-(n-q)
      = k[k(n-q)-1]-1-(n-q) = (k+1)[(k-1)(n-q)-1]$



      .Since $k geq 2$, we have $k+1 geq 3$. The last equality, whose left-hand side has
      positive integer divisors $1, 2, q$, and $2q$ only, implies that either $k+ 1 = q$
      or $k+1 = 2q$. If $k+1 = q$, then $(k-1)(n-q) = 3$, hence $(q-2)(n-q) = 3$.
      This leads to either $q-2 = 1$, $n-q = 3$, that is $q = 3, n = 6, k = q-1 = 2$,
      and, in view of $(1)$, $P = 5$, or else, $q-2 = 3$, $n-q = 1$, which gives $q = 5,
      n = 6, k = 4$, and in view of $(1)$, $p = 3$.
      On the other hand, if $k+1 = 2q$, then $(k-1)(n-q) = 2$, hence
      $2(q-1)(n-q) = 2$. This leads to $q-1 = 1$ and $n-q = 1$, or $q = 2, n = 3$,
      and, in view of $(1)$, $p = 2$. Thus, for positive integer $ n$, we have the
      following solutions in primes $p$ and $q$: $(p = q = 2, n = 3; 2)$,$ (p = 5,
      q = 3, n = 6)$, and $3$ ,$(p = 3, q = 5, n = 6)$. Only in the first solution all
      three numbers are primes.



      Note: If we denote by $displaystyle t_n = frac{n(n+1)}{2}$ the nth triangular number,
      then the equation $t_p+t_q = t_r$
      has only one solution in prime numbers, namely $p = q = 2, r = 3$.






      share|cite|improve this answer














      May this lead to a simple proof for your problem according to your unic example of solution . There is only one
      solution, namely $p = q = 2,r = 3$. To see that, we shall find all solutions
      of the equation $p(p+1)+q(q+1) = n(n+1)$ where $p$ and $q$ are primes and
      $n$ is a positive integer. Our equation yields
      $p(p+1) = n(n+1)-q(q+1) = (n-q)(n+q+1)$,
      and we must have $n > q$. Since $p$ is a prime, we have either $p|n-q$ or
      $p|n+q+1$. If $p|n-q$, then we have $pleq n-q$, which implies $p(p+1)
      leq (n-q)(n-q+1)$, and therefore $n+q+1 leq n-q+1$, which is impossible.
      Thus we have $p|n+q+ 1$, which means that for some positive integer $k$
      ,$n+q+1 = kp$, which implies $p+1 = k(n-q)tag1$.



      If we had $k = 1$, then $n+q+ 1 = p$ and $p+ 1 = n-q$, which gives $p-q
      = n+ 1$ and $p+q = n- 1$, which is impossible, because $(p+q)-(p-q)=2q>0$ and $(n-1)-(n+1)=-2<0$. Thus, $k > 1$. From $(1)$ we
      easily obtain:



      $2q = (n+q)-(n-q) = kp-1-(n-q)
      = k[k(n-q)-1]-1-(n-q) = (k+1)[(k-1)(n-q)-1]$



      .Since $k geq 2$, we have $k+1 geq 3$. The last equality, whose left-hand side has
      positive integer divisors $1, 2, q$, and $2q$ only, implies that either $k+ 1 = q$
      or $k+1 = 2q$. If $k+1 = q$, then $(k-1)(n-q) = 3$, hence $(q-2)(n-q) = 3$.
      This leads to either $q-2 = 1$, $n-q = 3$, that is $q = 3, n = 6, k = q-1 = 2$,
      and, in view of $(1)$, $P = 5$, or else, $q-2 = 3$, $n-q = 1$, which gives $q = 5,
      n = 6, k = 4$, and in view of $(1)$, $p = 3$.
      On the other hand, if $k+1 = 2q$, then $(k-1)(n-q) = 2$, hence
      $2(q-1)(n-q) = 2$. This leads to $q-1 = 1$ and $n-q = 1$, or $q = 2, n = 3$,
      and, in view of $(1)$, $p = 2$. Thus, for positive integer $ n$, we have the
      following solutions in primes $p$ and $q$: $(p = q = 2, n = 3; 2)$,$ (p = 5,
      q = 3, n = 6)$, and $3$ ,$(p = 3, q = 5, n = 6)$. Only in the first solution all
      three numbers are primes.



      Note: If we denote by $displaystyle t_n = frac{n(n+1)}{2}$ the nth triangular number,
      then the equation $t_p+t_q = t_r$
      has only one solution in prime numbers, namely $p = q = 2, r = 3$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 12 '16 at 23:23









      user236182

      11.9k11233




      11.9k11233










      answered Dec 12 '16 at 22:46









      zeraoulia rafik

      2,30611029




      2,30611029












      • It should be $p+q=n-1$.
        – user236182
        Dec 12 '16 at 22:51










      • It's still impossible because $(p+q)-(p-q)=2q>0$ while $(n-1)-(n+1)=-2<0$.
        – user236182
        Dec 12 '16 at 22:54










      • for k=1 it's impossible and i cited your case p+q=n-1 in the side of n+q=p-1
        – zeraoulia rafik
        Dec 12 '16 at 22:55










      • If $p+1=n-q$, then $p+q=n-1$.
        – user236182
        Dec 12 '16 at 22:57










      • And if $n+q+1=p$, then $p-q=n+1$.
        – user236182
        Dec 12 '16 at 22:58


















      • It should be $p+q=n-1$.
        – user236182
        Dec 12 '16 at 22:51










      • It's still impossible because $(p+q)-(p-q)=2q>0$ while $(n-1)-(n+1)=-2<0$.
        – user236182
        Dec 12 '16 at 22:54










      • for k=1 it's impossible and i cited your case p+q=n-1 in the side of n+q=p-1
        – zeraoulia rafik
        Dec 12 '16 at 22:55










      • If $p+1=n-q$, then $p+q=n-1$.
        – user236182
        Dec 12 '16 at 22:57










      • And if $n+q+1=p$, then $p-q=n+1$.
        – user236182
        Dec 12 '16 at 22:58
















      It should be $p+q=n-1$.
      – user236182
      Dec 12 '16 at 22:51




      It should be $p+q=n-1$.
      – user236182
      Dec 12 '16 at 22:51












      It's still impossible because $(p+q)-(p-q)=2q>0$ while $(n-1)-(n+1)=-2<0$.
      – user236182
      Dec 12 '16 at 22:54




      It's still impossible because $(p+q)-(p-q)=2q>0$ while $(n-1)-(n+1)=-2<0$.
      – user236182
      Dec 12 '16 at 22:54












      for k=1 it's impossible and i cited your case p+q=n-1 in the side of n+q=p-1
      – zeraoulia rafik
      Dec 12 '16 at 22:55




      for k=1 it's impossible and i cited your case p+q=n-1 in the side of n+q=p-1
      – zeraoulia rafik
      Dec 12 '16 at 22:55












      If $p+1=n-q$, then $p+q=n-1$.
      – user236182
      Dec 12 '16 at 22:57




      If $p+1=n-q$, then $p+q=n-1$.
      – user236182
      Dec 12 '16 at 22:57












      And if $n+q+1=p$, then $p-q=n+1$.
      – user236182
      Dec 12 '16 at 22:58




      And if $n+q+1=p$, then $p-q=n+1$.
      – user236182
      Dec 12 '16 at 22:58










      up vote
      0
      down vote













      Case $pne q$: We can assume that $p>q$, then $$r^2+r leq (p-1)^2+p^2+p-1+p=2p^2implies boxed{rleq psqrt{2}}$$



      From:



      $$p(p+1) =r^2- q^2+r-q =(r-q)(r+q+1)$$
      we get



      If $pmid r-q$ then $r+q+1mid p+1$, so $pleq r-q<r$ and $r+q+1leq p+1 implies r<p$ a contradiction.



      If $pmid r+q+1$ then $r-qmid p+1$. Since $r+q+1=kp$ for some integer $kgeq 1$ we have $$kpleq psqrt{2}+(p-1)+1 implies kleq 2$$



      Case 1: $k=2$ we get $r+q+1=2p$ and $p+1=2r-2q$ from where we get $3p=11$, no good.



      Case 2: $k=1$ we get $r+q+1=p$ and $p+1=r-q$ so $q=-1$ again contradiction.





      So $p=q$ and now we have to solve $$2p^2+2p = r^2+r$$



      So $$2p(p+1)=r(r+1)implies rmid 2p(p+1)$$



      If $rmid 2$ then $r=2$ which is impossibile.



      If $rmid p$ then $rleq p$ which is impossibile.



      If $rmid p+1$ then $rleq p+1$ but then $r=p+1$ since $r>p$, so $r=3$ and $p=2$.






      share|cite|improve this answer

























        up vote
        0
        down vote













        Case $pne q$: We can assume that $p>q$, then $$r^2+r leq (p-1)^2+p^2+p-1+p=2p^2implies boxed{rleq psqrt{2}}$$



        From:



        $$p(p+1) =r^2- q^2+r-q =(r-q)(r+q+1)$$
        we get



        If $pmid r-q$ then $r+q+1mid p+1$, so $pleq r-q<r$ and $r+q+1leq p+1 implies r<p$ a contradiction.



        If $pmid r+q+1$ then $r-qmid p+1$. Since $r+q+1=kp$ for some integer $kgeq 1$ we have $$kpleq psqrt{2}+(p-1)+1 implies kleq 2$$



        Case 1: $k=2$ we get $r+q+1=2p$ and $p+1=2r-2q$ from where we get $3p=11$, no good.



        Case 2: $k=1$ we get $r+q+1=p$ and $p+1=r-q$ so $q=-1$ again contradiction.





        So $p=q$ and now we have to solve $$2p^2+2p = r^2+r$$



        So $$2p(p+1)=r(r+1)implies rmid 2p(p+1)$$



        If $rmid 2$ then $r=2$ which is impossibile.



        If $rmid p$ then $rleq p$ which is impossibile.



        If $rmid p+1$ then $rleq p+1$ but then $r=p+1$ since $r>p$, so $r=3$ and $p=2$.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Case $pne q$: We can assume that $p>q$, then $$r^2+r leq (p-1)^2+p^2+p-1+p=2p^2implies boxed{rleq psqrt{2}}$$



          From:



          $$p(p+1) =r^2- q^2+r-q =(r-q)(r+q+1)$$
          we get



          If $pmid r-q$ then $r+q+1mid p+1$, so $pleq r-q<r$ and $r+q+1leq p+1 implies r<p$ a contradiction.



          If $pmid r+q+1$ then $r-qmid p+1$. Since $r+q+1=kp$ for some integer $kgeq 1$ we have $$kpleq psqrt{2}+(p-1)+1 implies kleq 2$$



          Case 1: $k=2$ we get $r+q+1=2p$ and $p+1=2r-2q$ from where we get $3p=11$, no good.



          Case 2: $k=1$ we get $r+q+1=p$ and $p+1=r-q$ so $q=-1$ again contradiction.





          So $p=q$ and now we have to solve $$2p^2+2p = r^2+r$$



          So $$2p(p+1)=r(r+1)implies rmid 2p(p+1)$$



          If $rmid 2$ then $r=2$ which is impossibile.



          If $rmid p$ then $rleq p$ which is impossibile.



          If $rmid p+1$ then $rleq p+1$ but then $r=p+1$ since $r>p$, so $r=3$ and $p=2$.






          share|cite|improve this answer












          Case $pne q$: We can assume that $p>q$, then $$r^2+r leq (p-1)^2+p^2+p-1+p=2p^2implies boxed{rleq psqrt{2}}$$



          From:



          $$p(p+1) =r^2- q^2+r-q =(r-q)(r+q+1)$$
          we get



          If $pmid r-q$ then $r+q+1mid p+1$, so $pleq r-q<r$ and $r+q+1leq p+1 implies r<p$ a contradiction.



          If $pmid r+q+1$ then $r-qmid p+1$. Since $r+q+1=kp$ for some integer $kgeq 1$ we have $$kpleq psqrt{2}+(p-1)+1 implies kleq 2$$



          Case 1: $k=2$ we get $r+q+1=2p$ and $p+1=2r-2q$ from where we get $3p=11$, no good.



          Case 2: $k=1$ we get $r+q+1=p$ and $p+1=r-q$ so $q=-1$ again contradiction.





          So $p=q$ and now we have to solve $$2p^2+2p = r^2+r$$



          So $$2p(p+1)=r(r+1)implies rmid 2p(p+1)$$



          If $rmid 2$ then $r=2$ which is impossibile.



          If $rmid p$ then $rleq p$ which is impossibile.



          If $rmid p+1$ then $rleq p+1$ but then $r=p+1$ since $r>p$, so $r=3$ and $p=2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 14 at 23:13









          greedoid

          34.4k114488




          34.4k114488






















              up vote
              0
              down vote













              From the given equation
              $$p(p+1)+q(q+1)=r(r+1)$$
              it follows that $p < r$ and $q < r$.



              Next, another inequality which will be useful later . . .



              Claim:$;p+q > r$.



              Proof:
              begin{align*}
              &p(p+1)+q(q+1)=r(r+1)\[4pt]
              implies;&(p^2+q^2)+(p+q)=r(r+1)\[4pt]
              implies;&(p+q)^2+(p+q)>r(r+1)\[4pt]
              implies;&(p+q)(p+q+1)>r(r+1)\[4pt]
              implies;&p+q>r\[4pt]
              end{align*}

              as claimed.



              Returning to the main problem . . .



              First suppose $p=q$.



              Then the given equation reduces to
              $$2p(p+1)=r(r+1)$$
              hence, since $r > p$, it follows that $r|(p+1)$.



              But then $p < r le p+1$, so $r=p+1$, which implies $p=2$, and $r=3$.



              It can be verified that the triple $(p,q,r)=(2,2,3)$ satisfies the given equation.



              Next suppose $p,q$ are distinct.



              Without loss of generality, assume $p < q$.



              Suppose $;p=2$.



              Then from $p < q < r$, we get $qge 3$ and $rge 5$, hence
              begin{align*}
              &p(p+1)+q(q+1)=r(r+1)\[4pt]
              implies;&(2)(2+1)+q(q+1)=r(r+1)\[4pt]
              implies;&r(r+1)-q(q+1)=6\[4pt]
              implies;&(r-q)(q+r+1)=6\[4pt]
              implies;&(q+r+1)mid 6\[4pt]
              implies;&q+r+1le 6\[4pt]
              end{align*}

              contradiction, since $q+r+1ge 3+5+1=9$.



              Hence we must have $p > 2$.



              Since $p+q > r$, it follows that $pnotmid r-q$, and $qnotmid r-p$.
              begin{align*}
              text{Then};;&p(p+1)+q(q+1)=r(r+1)\[4pt]
              implies;&p(p+1)=r(r+1)-q(q+1)\[4pt]
              implies;&p(p+1)=(q+r+1)(r-q)\[4pt]
              implies;&pmid (q+r+1)\[4pt]
              implies;&pmid (p+q+r+1)\[12pt]
              text{and};,&p(p+1)+q(q+1)=r(r+1)\[4pt]
              implies;&q(q+1)=r(r+1)-p(p+1)\[4pt]
              implies;&q(q+1)=(p+r+1)(r-p)\[4pt]
              implies;&qmid (p+r+1)\[4pt]
              implies;&qmid (p+q+r+1)\[12pt]
              text{hence};,&pqmid (p+q+r+1)\[4pt]
              implies;&pqle p+q+r+1\[4pt]
              implies;&pq < p+q+(p+q)+1\[4pt]
              implies;&pq-2p-2q < 1\[4pt]
              implies;&(p-2)(q-2) < 5\[4pt]
              implies;&q-2 < 5\[4pt]
              implies;&q < 7\[4pt]
              implies;&qle 5\[4pt]
              implies;&(p,q)=(3,5)\[4pt]
              implies;&(3)(3+1)+(5)(5+1)=r(r+1)\[4pt]
              implies;&r=6\[4pt]
              end{align*}

              contradiction, since $6$ is not prime.



              Therefore the only solution is $(p,q,r)=(2,2,3)$.






              share|cite|improve this answer



























                up vote
                0
                down vote













                From the given equation
                $$p(p+1)+q(q+1)=r(r+1)$$
                it follows that $p < r$ and $q < r$.



                Next, another inequality which will be useful later . . .



                Claim:$;p+q > r$.



                Proof:
                begin{align*}
                &p(p+1)+q(q+1)=r(r+1)\[4pt]
                implies;&(p^2+q^2)+(p+q)=r(r+1)\[4pt]
                implies;&(p+q)^2+(p+q)>r(r+1)\[4pt]
                implies;&(p+q)(p+q+1)>r(r+1)\[4pt]
                implies;&p+q>r\[4pt]
                end{align*}

                as claimed.



                Returning to the main problem . . .



                First suppose $p=q$.



                Then the given equation reduces to
                $$2p(p+1)=r(r+1)$$
                hence, since $r > p$, it follows that $r|(p+1)$.



                But then $p < r le p+1$, so $r=p+1$, which implies $p=2$, and $r=3$.



                It can be verified that the triple $(p,q,r)=(2,2,3)$ satisfies the given equation.



                Next suppose $p,q$ are distinct.



                Without loss of generality, assume $p < q$.



                Suppose $;p=2$.



                Then from $p < q < r$, we get $qge 3$ and $rge 5$, hence
                begin{align*}
                &p(p+1)+q(q+1)=r(r+1)\[4pt]
                implies;&(2)(2+1)+q(q+1)=r(r+1)\[4pt]
                implies;&r(r+1)-q(q+1)=6\[4pt]
                implies;&(r-q)(q+r+1)=6\[4pt]
                implies;&(q+r+1)mid 6\[4pt]
                implies;&q+r+1le 6\[4pt]
                end{align*}

                contradiction, since $q+r+1ge 3+5+1=9$.



                Hence we must have $p > 2$.



                Since $p+q > r$, it follows that $pnotmid r-q$, and $qnotmid r-p$.
                begin{align*}
                text{Then};;&p(p+1)+q(q+1)=r(r+1)\[4pt]
                implies;&p(p+1)=r(r+1)-q(q+1)\[4pt]
                implies;&p(p+1)=(q+r+1)(r-q)\[4pt]
                implies;&pmid (q+r+1)\[4pt]
                implies;&pmid (p+q+r+1)\[12pt]
                text{and};,&p(p+1)+q(q+1)=r(r+1)\[4pt]
                implies;&q(q+1)=r(r+1)-p(p+1)\[4pt]
                implies;&q(q+1)=(p+r+1)(r-p)\[4pt]
                implies;&qmid (p+r+1)\[4pt]
                implies;&qmid (p+q+r+1)\[12pt]
                text{hence};,&pqmid (p+q+r+1)\[4pt]
                implies;&pqle p+q+r+1\[4pt]
                implies;&pq < p+q+(p+q)+1\[4pt]
                implies;&pq-2p-2q < 1\[4pt]
                implies;&(p-2)(q-2) < 5\[4pt]
                implies;&q-2 < 5\[4pt]
                implies;&q < 7\[4pt]
                implies;&qle 5\[4pt]
                implies;&(p,q)=(3,5)\[4pt]
                implies;&(3)(3+1)+(5)(5+1)=r(r+1)\[4pt]
                implies;&r=6\[4pt]
                end{align*}

                contradiction, since $6$ is not prime.



                Therefore the only solution is $(p,q,r)=(2,2,3)$.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  From the given equation
                  $$p(p+1)+q(q+1)=r(r+1)$$
                  it follows that $p < r$ and $q < r$.



                  Next, another inequality which will be useful later . . .



                  Claim:$;p+q > r$.



                  Proof:
                  begin{align*}
                  &p(p+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&(p^2+q^2)+(p+q)=r(r+1)\[4pt]
                  implies;&(p+q)^2+(p+q)>r(r+1)\[4pt]
                  implies;&(p+q)(p+q+1)>r(r+1)\[4pt]
                  implies;&p+q>r\[4pt]
                  end{align*}

                  as claimed.



                  Returning to the main problem . . .



                  First suppose $p=q$.



                  Then the given equation reduces to
                  $$2p(p+1)=r(r+1)$$
                  hence, since $r > p$, it follows that $r|(p+1)$.



                  But then $p < r le p+1$, so $r=p+1$, which implies $p=2$, and $r=3$.



                  It can be verified that the triple $(p,q,r)=(2,2,3)$ satisfies the given equation.



                  Next suppose $p,q$ are distinct.



                  Without loss of generality, assume $p < q$.



                  Suppose $;p=2$.



                  Then from $p < q < r$, we get $qge 3$ and $rge 5$, hence
                  begin{align*}
                  &p(p+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&(2)(2+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&r(r+1)-q(q+1)=6\[4pt]
                  implies;&(r-q)(q+r+1)=6\[4pt]
                  implies;&(q+r+1)mid 6\[4pt]
                  implies;&q+r+1le 6\[4pt]
                  end{align*}

                  contradiction, since $q+r+1ge 3+5+1=9$.



                  Hence we must have $p > 2$.



                  Since $p+q > r$, it follows that $pnotmid r-q$, and $qnotmid r-p$.
                  begin{align*}
                  text{Then};;&p(p+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&p(p+1)=r(r+1)-q(q+1)\[4pt]
                  implies;&p(p+1)=(q+r+1)(r-q)\[4pt]
                  implies;&pmid (q+r+1)\[4pt]
                  implies;&pmid (p+q+r+1)\[12pt]
                  text{and};,&p(p+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&q(q+1)=r(r+1)-p(p+1)\[4pt]
                  implies;&q(q+1)=(p+r+1)(r-p)\[4pt]
                  implies;&qmid (p+r+1)\[4pt]
                  implies;&qmid (p+q+r+1)\[12pt]
                  text{hence};,&pqmid (p+q+r+1)\[4pt]
                  implies;&pqle p+q+r+1\[4pt]
                  implies;&pq < p+q+(p+q)+1\[4pt]
                  implies;&pq-2p-2q < 1\[4pt]
                  implies;&(p-2)(q-2) < 5\[4pt]
                  implies;&q-2 < 5\[4pt]
                  implies;&q < 7\[4pt]
                  implies;&qle 5\[4pt]
                  implies;&(p,q)=(3,5)\[4pt]
                  implies;&(3)(3+1)+(5)(5+1)=r(r+1)\[4pt]
                  implies;&r=6\[4pt]
                  end{align*}

                  contradiction, since $6$ is not prime.



                  Therefore the only solution is $(p,q,r)=(2,2,3)$.






                  share|cite|improve this answer














                  From the given equation
                  $$p(p+1)+q(q+1)=r(r+1)$$
                  it follows that $p < r$ and $q < r$.



                  Next, another inequality which will be useful later . . .



                  Claim:$;p+q > r$.



                  Proof:
                  begin{align*}
                  &p(p+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&(p^2+q^2)+(p+q)=r(r+1)\[4pt]
                  implies;&(p+q)^2+(p+q)>r(r+1)\[4pt]
                  implies;&(p+q)(p+q+1)>r(r+1)\[4pt]
                  implies;&p+q>r\[4pt]
                  end{align*}

                  as claimed.



                  Returning to the main problem . . .



                  First suppose $p=q$.



                  Then the given equation reduces to
                  $$2p(p+1)=r(r+1)$$
                  hence, since $r > p$, it follows that $r|(p+1)$.



                  But then $p < r le p+1$, so $r=p+1$, which implies $p=2$, and $r=3$.



                  It can be verified that the triple $(p,q,r)=(2,2,3)$ satisfies the given equation.



                  Next suppose $p,q$ are distinct.



                  Without loss of generality, assume $p < q$.



                  Suppose $;p=2$.



                  Then from $p < q < r$, we get $qge 3$ and $rge 5$, hence
                  begin{align*}
                  &p(p+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&(2)(2+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&r(r+1)-q(q+1)=6\[4pt]
                  implies;&(r-q)(q+r+1)=6\[4pt]
                  implies;&(q+r+1)mid 6\[4pt]
                  implies;&q+r+1le 6\[4pt]
                  end{align*}

                  contradiction, since $q+r+1ge 3+5+1=9$.



                  Hence we must have $p > 2$.



                  Since $p+q > r$, it follows that $pnotmid r-q$, and $qnotmid r-p$.
                  begin{align*}
                  text{Then};;&p(p+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&p(p+1)=r(r+1)-q(q+1)\[4pt]
                  implies;&p(p+1)=(q+r+1)(r-q)\[4pt]
                  implies;&pmid (q+r+1)\[4pt]
                  implies;&pmid (p+q+r+1)\[12pt]
                  text{and};,&p(p+1)+q(q+1)=r(r+1)\[4pt]
                  implies;&q(q+1)=r(r+1)-p(p+1)\[4pt]
                  implies;&q(q+1)=(p+r+1)(r-p)\[4pt]
                  implies;&qmid (p+r+1)\[4pt]
                  implies;&qmid (p+q+r+1)\[12pt]
                  text{hence};,&pqmid (p+q+r+1)\[4pt]
                  implies;&pqle p+q+r+1\[4pt]
                  implies;&pq < p+q+(p+q)+1\[4pt]
                  implies;&pq-2p-2q < 1\[4pt]
                  implies;&(p-2)(q-2) < 5\[4pt]
                  implies;&q-2 < 5\[4pt]
                  implies;&q < 7\[4pt]
                  implies;&qle 5\[4pt]
                  implies;&(p,q)=(3,5)\[4pt]
                  implies;&(3)(3+1)+(5)(5+1)=r(r+1)\[4pt]
                  implies;&r=6\[4pt]
                  end{align*}

                  contradiction, since $6$ is not prime.



                  Therefore the only solution is $(p,q,r)=(2,2,3)$.







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                  edited Nov 14 at 23:52

























                  answered Nov 14 at 23:37









                  quasi

                  35.9k22562




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