<s>[INST] For how many integers $(3n)^4-(n-10)^4$ is a perfect square? 
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$$
Is this getting simpler or harder this way? Do you have a suggestion?   
 [/INST] $$(3n)^4-(n-10)^4=m^2\Rightarrow\begin{cases} 9n^2=x^2+y^2\\n^2-20n+100=x^2-y^2\text{ or}\space 2xy\end{cases}$$ WLG we can consider $n,x,y$ without common factors then $n$ cannot be even because the well known parametrization of  Pythagorean triples we used above.
$$n\text{ odd}\Rightarrow 10n^2-20n+100=2x^2\Rightarrow 5n^2-10n+50=x^2$$
Then $$5n^2-10n+50=25x_1^2\Rightarrow n^2-2n+10=5x_1^2\iff(n-1)^2+3^2=5x_1^2$$ This implies $5x_1^2=25$ which is absurde.
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