KöMaL (Metresis) 1894, Problem 12

Problem:
Find a four digit number which is a perfect square, knowing that the first two digit number exceeds by $1$ the last two digit number.

Solution:
Let $x$ be the first two digit number and let $y^2$ be the four digit number. Then, $31 < y < 100$ and
$$100(x+1)+x=y^2 \iff 101x=(y-10)(y+10).$$ So, $101$ divides one between $y-10$ and $y+10$. But $y-10<101$, so $101|(y+10)$. Since $41<y+10<110$, $y+10=101$, which gives $y=91$. Therefore, $y^2=8281$.

KöMaL (Metresis) 1894, Problem 10

Problem:
Find all integers such that their fifth power minus three times their square is equal to $216$.

Solution:
Suppose that $n$ is an integer such that $n^5-3n^2=216=2^3\cdot3^3$. Then $$n^2(n^3-3)=216.$$ From this equality we get $n>1$. Let $d=\gcd(n^2,n^3-3)$. Then $d=1,3$. If $d=1$, then both $n^2$ and $(n^3-3)$ are perfect cubes, impossible since $$(n-1)^3 < n^3-3 < n^3.$$ If $d=3$, then $3^2|n^2$, so $n^2=9a, n^3-3=3b$ and $\gcd(a,b)=1$. Therefore, $ab=8$ and $b+1=3^{3k-1}$ for some $k \in \mathbb{N}$, which gives $a=1,b=8$, i.e. $n=3$.

Note. It was faster to note that $n^2$ and $(n^3-3)$ have different parity, so one of the two is odd. If $n^2$ is odd, $n^2=9$ which yields $n=3$, if $(n^3-3)$ is odd, then $n^3-3=1,3,9,27$, i.e. no solution.