Subject (3). Determine the natural numbers $m, n$ such that $85^{m}-n^{4}=4$.
Solution. The Sophie Germain factorization gives $85^{m}=n^{4}+4=\left(n^{4}+4 n^{2}+4\right)-$ $4 n^{2}=\left(n^{2}+2\right)^{2}-(2 n)^{2}=\left(n^{2}-2 n+2\right)\left(n^{2}+2 n+2\right)$. Since we need $n$ to be odd, it follows that $\operatorname{gcd}\left(n^{2}-2 n+2, n^{2}+2 n+2\right)=1$, so $n^{2}-2 n+2=1$, $n^{2}+2 n+2=85^{m}$, or $n^{2}-2 n+2=5^{m}, n^{2}+2 n+2=17^{m}$.

The first case is impossible, as it requires $n=1$, which does not verify. The second case leads to $(n-1)^{2}=5^{m}-11$ we have $5^{m}-1<\left(3^{m}\right)^{2}<\left(4^{m}\right)^{2}<17^{m}-1$, so we must have $m=1, n=3$, which verifies $85-3^{4}=4$ (clearly $m=0$ does not verify).