On the factorial equations $A! B! =C!$ and $A!B!C! = D!$












14














I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient $binom{10}{3}$. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything
$$
frac {8cdot9cdot10}{2cdot3} = 120
$$
And then it hit me that $8cdot9cdot10 = 6!$ and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this
$$
a!b = prod_{n=b}^an = frac{a!}{(b-1)!},quad a, b in mathbb{Z}^+, quad age b
$$
so that $a! = a!1$ (the notation was invented just now, and inspired by the $nCr$-notation for binomial coefficients). Now, apart from the trivial examples $(n!)!(n!) = n!$ and $a!1 = a!2 = a!$, when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, $10!8$ is both.










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  • 2




    You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
    – Henry
    Feb 24 '12 at 0:41










  • You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
    – Henry
    Feb 24 '12 at 1:05










  • As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
    – Arthur
    Feb 24 '12 at 13:24






  • 4




    The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
    – John Bentin
    Sep 15 '13 at 12:56










  • Similar question was answered at MathOverflow.
    – Alexey Ustinov
    Mar 30 '18 at 11:47
















14














I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient $binom{10}{3}$. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything
$$
frac {8cdot9cdot10}{2cdot3} = 120
$$
And then it hit me that $8cdot9cdot10 = 6!$ and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this
$$
a!b = prod_{n=b}^an = frac{a!}{(b-1)!},quad a, b in mathbb{Z}^+, quad age b
$$
so that $a! = a!1$ (the notation was invented just now, and inspired by the $nCr$-notation for binomial coefficients). Now, apart from the trivial examples $(n!)!(n!) = n!$ and $a!1 = a!2 = a!$, when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, $10!8$ is both.










share|cite|improve this question




















  • 2




    You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
    – Henry
    Feb 24 '12 at 0:41










  • You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
    – Henry
    Feb 24 '12 at 1:05










  • As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
    – Arthur
    Feb 24 '12 at 13:24






  • 4




    The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
    – John Bentin
    Sep 15 '13 at 12:56










  • Similar question was answered at MathOverflow.
    – Alexey Ustinov
    Mar 30 '18 at 11:47














14












14








14


9





I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient $binom{10}{3}$. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything
$$
frac {8cdot9cdot10}{2cdot3} = 120
$$
And then it hit me that $8cdot9cdot10 = 6!$ and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this
$$
a!b = prod_{n=b}^an = frac{a!}{(b-1)!},quad a, b in mathbb{Z}^+, quad age b
$$
so that $a! = a!1$ (the notation was invented just now, and inspired by the $nCr$-notation for binomial coefficients). Now, apart from the trivial examples $(n!)!(n!) = n!$ and $a!1 = a!2 = a!$, when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, $10!8$ is both.










share|cite|improve this question















I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient $binom{10}{3}$. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything
$$
frac {8cdot9cdot10}{2cdot3} = 120
$$
And then it hit me that $8cdot9cdot10 = 6!$ and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this
$$
a!b = prod_{n=b}^an = frac{a!}{(b-1)!},quad a, b in mathbb{Z}^+, quad age b
$$
so that $a! = a!1$ (the notation was invented just now, and inspired by the $nCr$-notation for binomial coefficients). Now, apart from the trivial examples $(n!)!(n!) = n!$ and $a!1 = a!2 = a!$, when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, $10!8$ is both.







elementary-number-theory factorial






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edited Feb 24 '12 at 2:00









Aryabhata

70.1k6156246




70.1k6156246










asked Feb 24 '12 at 0:23









ArthurArthur

111k7105186




111k7105186








  • 2




    You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
    – Henry
    Feb 24 '12 at 0:41










  • You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
    – Henry
    Feb 24 '12 at 1:05










  • As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
    – Arthur
    Feb 24 '12 at 13:24






  • 4




    The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
    – John Bentin
    Sep 15 '13 at 12:56










  • Similar question was answered at MathOverflow.
    – Alexey Ustinov
    Mar 30 '18 at 11:47














  • 2




    You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
    – Henry
    Feb 24 '12 at 0:41










  • You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
    – Henry
    Feb 24 '12 at 1:05










  • As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
    – Arthur
    Feb 24 '12 at 13:24






  • 4




    The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
    – John Bentin
    Sep 15 '13 at 12:56










  • Similar question was answered at MathOverflow.
    – Alexey Ustinov
    Mar 30 '18 at 11:47








2




2




You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
– Henry
Feb 24 '12 at 0:41




You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
– Henry
Feb 24 '12 at 0:41












You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
– Henry
Feb 24 '12 at 1:05




You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
– Henry
Feb 24 '12 at 1:05












As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
– Arthur
Feb 24 '12 at 13:24




As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
– Arthur
Feb 24 '12 at 13:24




4




4




The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
– John Bentin
Sep 15 '13 at 12:56




The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
– John Bentin
Sep 15 '13 at 12:56












Similar question was answered at MathOverflow.
– Alexey Ustinov
Mar 30 '18 at 11:47




Similar question was answered at MathOverflow.
– Alexey Ustinov
Mar 30 '18 at 11:47










3 Answers
3






active

oldest

votes


















22














See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).






share|cite|improve this answer





















  • Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
    – Arthur
    Feb 24 '12 at 13:25



















10














Immediately we see that any $n$ that's a power of 2 should work. e.g.



$4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$



Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.



However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by



$n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:



$n(n-1)(n-2) ... (n-k) = P(n) = A!$



Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf






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  • Very interesting reference, thank you!
    – Jose Brox
    Feb 18 '16 at 20:43



















2














Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.



Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.



Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
$$
P(n(n+1)) > 4 log n
$$
would imply that the general equation $(1)$ has finitely many non-trivial solutions.



The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.



More references:




  1. Discussion on ArtofProblemSolving

  2. Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.

  3. Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.






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    3 Answers
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    3 Answers
    3






    active

    oldest

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    active

    oldest

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    active

    oldest

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    22














    See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).






    share|cite|improve this answer





















    • Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
      – Arthur
      Feb 24 '12 at 13:25
















    22














    See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).






    share|cite|improve this answer





















    • Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
      – Arthur
      Feb 24 '12 at 13:25














    22












    22








    22






    See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).






    share|cite|improve this answer












    See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Feb 24 '12 at 1:40









    Gerry MyersonGerry Myerson

    146k8147298




    146k8147298












    • Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
      – Arthur
      Feb 24 '12 at 13:25


















    • Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
      – Arthur
      Feb 24 '12 at 13:25
















    Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
    – Arthur
    Feb 24 '12 at 13:25




    Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
    – Arthur
    Feb 24 '12 at 13:25











    10














    Immediately we see that any $n$ that's a power of 2 should work. e.g.



    $4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$



    Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.



    However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by



    $n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:



    $n(n-1)(n-2) ... (n-k) = P(n) = A!$



    Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf






    share|cite|improve this answer























    • Very interesting reference, thank you!
      – Jose Brox
      Feb 18 '16 at 20:43
















    10














    Immediately we see that any $n$ that's a power of 2 should work. e.g.



    $4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$



    Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.



    However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by



    $n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:



    $n(n-1)(n-2) ... (n-k) = P(n) = A!$



    Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf






    share|cite|improve this answer























    • Very interesting reference, thank you!
      – Jose Brox
      Feb 18 '16 at 20:43














    10












    10








    10






    Immediately we see that any $n$ that's a power of 2 should work. e.g.



    $4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$



    Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.



    However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by



    $n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:



    $n(n-1)(n-2) ... (n-k) = P(n) = A!$



    Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf






    share|cite|improve this answer














    Immediately we see that any $n$ that's a power of 2 should work. e.g.



    $4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$



    Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.



    However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by



    $n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:



    $n(n-1)(n-2) ... (n-k) = P(n) = A!$



    Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 5 '18 at 23:41

























    answered Feb 11 '14 at 0:12









    Alex MeiburgAlex Meiburg

    1,805517




    1,805517












    • Very interesting reference, thank you!
      – Jose Brox
      Feb 18 '16 at 20:43


















    • Very interesting reference, thank you!
      – Jose Brox
      Feb 18 '16 at 20:43
















    Very interesting reference, thank you!
    – Jose Brox
    Feb 18 '16 at 20:43




    Very interesting reference, thank you!
    – Jose Brox
    Feb 18 '16 at 20:43











    2














    Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
    People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.



    Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
    For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.



    Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
    $$
    P(n(n+1)) > 4 log n
    $$
    would imply that the general equation $(1)$ has finitely many non-trivial solutions.



    The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.



    More references:




    1. Discussion on ArtofProblemSolving

    2. Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.

    3. Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.






    share|cite|improve this answer




























      2














      Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
      People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.



      Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
      For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.



      Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
      $$
      P(n(n+1)) > 4 log n
      $$
      would imply that the general equation $(1)$ has finitely many non-trivial solutions.



      The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.



      More references:




      1. Discussion on ArtofProblemSolving

      2. Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.

      3. Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.






      share|cite|improve this answer


























        2












        2








        2






        Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
        People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.



        Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
        For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.



        Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
        $$
        P(n(n+1)) > 4 log n
        $$
        would imply that the general equation $(1)$ has finitely many non-trivial solutions.



        The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.



        More references:




        1. Discussion on ArtofProblemSolving

        2. Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.

        3. Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.






        share|cite|improve this answer














        Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
        People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.



        Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
        For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.



        Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
        $$
        P(n(n+1)) > 4 log n
        $$
        would imply that the general equation $(1)$ has finitely many non-trivial solutions.



        The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.



        More references:




        1. Discussion on ArtofProblemSolving

        2. Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.

        3. Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 5 '18 at 9:21

























        answered May 5 '18 at 8:37









        Amir HosseinAmir Hossein

        2,38111650




        2,38111650






























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