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Consider the Diophantine equation $$m(m-1)(m-2)(m-3) = 24(n^2 + 9)\,.$$ Prove that there are no integer solutions.

One way to show this has no integer solutions is by considering modulo $7$ (easy to verify with it).

I am curious whether there is a slightly less $``$random$``$ way to solve this problem such as using the fact that if $p\equiv 3 \pmod 4$ divides $x^2 + y^2$, then $p$ must divide both $x$ and $y$. This looks convenient since the left-hand side has a multiplier which is $\equiv 3 \pmod 4$ (and hence such a $p$ surely exists) and we will be done provided we can take $p\neq 3$ (since the only prime $p\equiv 3 \pmod 4$ which divides $y=3$ is $3$ itself). Any idea if this method could work?

I am of course also open to see other ideas. Any help appreciated!

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    $\begingroup$ In general, such equations are extremely difficult to solve. But over and over again other users surprise me with methods that can apparently handle a large number of cases. If you can show that there is no solution modulo some $m$, you are lucky. Usually, this is not the case. Or the object could be to find all solutions, if there is one, the modulo-approach alone cannot work. Equations of that form usually have only finite many integer solutions because of Falting's theorem, but I could nowhere find a handy criterion to decide when it applies. $\endgroup$
    – Peter
    Jul 1, 2020 at 15:16
  • $\begingroup$ I am not sure whether the negative result of Hilbert's tenth problem applies already in two variables or whether we can solve every such diophantine equation.completely. $\endgroup$
    – Peter
    Jul 1, 2020 at 15:19
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    $\begingroup$ Looking modulo $7$ is definitely not random. Let $s:=m^2-3m+1$. Then, $$s^2=24n^2+217\,.$$ Since $7$ divides $217=7\cdot 31$, we get $$s^2\equiv 24n^2\equiv 3n^2\pmod{7}\,.$$ Now, $3$ is not a quadratic residue modulo $7$, so the equivalence $s^2\equiv 3n^2\pmod{7}$ implies that both $s$ and $n$ are divisible by $7$. However, this means $7^2$ divides $217$, which is a contradiction. You can also use $31$ instead of $7$, and this still works because $24$ is a quadratic nonresidue modulo $31$. $\endgroup$ Jul 1, 2020 at 15:34
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    $\begingroup$ In general, when you have a product or sum of consecutive integers, it is a good idea to consider making the central number the variable, as it allows for using difference of squares in a product or cancellations in a sum. If the number of numbers is even, you can take the average of the middle two numbers. Take a look at the method that I posted below. $\endgroup$
    – Favst
    Jul 1, 2020 at 16:08

1 Answer 1

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Here is one approach that is more motivated. Let $k$ be the integer in between $m-1$ and $m-2.$ So $$k=\frac{(m-1)+(m-2)}{2}=m-\frac{3}{2}.$$ Then the left side of the equation is \begin{align*} m(m-1)(m-2)(m-3) &= \left(k+\frac{3}{2}\right)\left(k+\frac{1}{2}\right)\left(k-\frac{1}{2}\right)\left(k-\frac{3}{2}\right)\\ &=\left(k^2-\frac{9}{4}\right)\left(k^2-\frac{1}{4}\right)\\ &= k^4 -\frac{10}{4}k^2+\frac{9}{16}\\ &= k^4 -\frac{5}{2}k^2+\frac{25}{16}-\frac{25}{16}+\frac{9}{16}\\ &= \left(k^2 -\frac{5}{4}\right)^2 - 1\\ &= \left(\left(m-\frac{3}{2}\right)^2-\frac{5}{4}\right)^2-1\\ &= (m^2-3m+1)^2 -1. \end{align*} So the equation is $$(m^2-3m+1)^2 - 24n^2=24\cdot 9 +1 =7\cdot 31.$$ This is a good enough reason to try $\mod 7.$ Then $$(m^2-3m+1)^2\equiv 3n^2 \pmod{7}.$$ The quadratic residues modulo $7$ are $0,1,2,4.$ The only two residues where one is three times the other are $0$ and $0.$ So $m^2-3m+1$ and $n$ are both divisible by $7.$ In the first case $$m^2-3m+1\equiv 0\pmod 7.$$ Equivalently, $$(m+2)^2\equiv 3\pmod{7}.$$ But none of the quadratic residues are $3$ so that's a contradiction to the existence of a solution $(m,n).$

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