For how many integers $n$ the expression $$(3n)^4-(n-10)^4$$ becomes a perfect square?
$$$$One way is: Let $x=3n$ and $y=n-10$ to get $$x^4-y^4=z^2.$$ This equation does not have non-trivial solutions in integers (I do not want to discuss about this).
I was trying to solve this problem with simpler elementary methods. For example we can write the equation as $$40(n+5)(2n-5)(n^2-2n+10)=m^2.$$ Observe that $40 \mid m^2$. Thus $20 \mid m$. Write $m=20k$ to get $$(n+5)(2n-5)(n^2-2n+10)=10k^2,$$ Where $k$ is a non-negative integer. Look mod $5$. We get $n^3(n-2) \equiv 0 \pmod{5}$. So we have either $n \equiv 0 \pmod{5}$ or $n \equiv 2 \pmod{5}$.
If $n \equiv 0 \pmod{5}$ then let $n=5N$ to get $$125(N+1)(2N-1)(5N^2-2N+2)=10k^2$$ It follows that $25 \mid k^2$ and $k=5K$. Now we have the following equation $$(N+1)(2N-1)(5N^2-2N+2)=2K^2$$
Is this getting simpler or harder this way? Do you have a suggestion?