
Problem 1. Find all triples $(x, y, z)$ of natural numbers such that $y$ is a prime number, $y$ and 3 do not divide $z$, and $x^{3}-y^{3}=z^{2}$.

Nikolay Nikolov


Solution: Since $\left.(x-y)\left((x-y)^{2}+3 x y\right)\right)=x^{3}-y^{3}=z^{2}$, it follows from the conditions of the problem that $x-y$ and $(x-y)^{2}+3 x y$ are relatively prime. Therefore $x-y=u^{2}$ and $x^{2}+x y+y^{2}=v^{2}$. Thus $3 y^{2}=(2 v-2 x-y)(2 v+2 x+y)$ and since $y$ is a prime number, there are three cases to consider:

1. $2 v-2 x-y=y, 2 v+2 x+y=3 y$. Now $x=0$, which is impossible.
2. $2 v-2 x-y=1,2 v+2 x+y=3 y^{2}$. Now $3 y^{2}-1=2(2 x+$ $y)=2\left(2 u^{2}+3 y\right)$ and it follows that 3 divides $u^{2}+1$, which is impossible.
3. $2 v-2 x-y=3,2 v+2 x+y=y^{2}$. Now $y^{2}-3=2(2 x+y)=$ $2\left(2 u^{2}+3 y\right)$ and it follows that $(y-3)^{2}-(2 u)^{2}=12$. Therefore $y=7, u=1, x=8, z=13$. Direct verification shows that $(8,7,13)$ is a solution of the problem.
