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!
Best Answer
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 $(m,n).$