For how many integers $(3n)^4-(n-10)^4$ is a perfect square?

Question

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?

Answer 1

$$(3n)^4-(n-10)^4=m^2\Rightarrow\begin{cases} 9n^2=x^2+y^2\\n^2-20n+100=x^2-y^2\text{ or}\space 2xy\end{cases}$$ 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.

Answer 2

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.