<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$$
Is this getting simpler or harder this way? Do you have a suggestion?   
 [/INST] Using
$$\gcd(N+1,2N-1)=\gcd(N+1,-3)\mid 3$$
$$\gcd(N+1,5N^2-2N+2)=\gcd(N+1,-7N+2)=\gcd(N+1,9)\mid 9$$
$$\gcd(2N-1,5N^2-2N+2)=\gcd(2N-1,5)\mid 5 $$
we see that the factors are almost coprime, hence must almost (i.e., up to small factors involving at most a $2$, a $3$, and/or a $5$) be square.
Then from $\frac{2N-1}{N+1}\approx 2$ you might arrive at a contradiction.
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