Estonian Math Competitions

Answer: $(0,1,0),(-1,0,0)$.

Substituting $x=a-b$, $y=a+b$ we obtain the equation $$x^3{y}^2=c^2+2x+1,$$ which suffices to be solved in integers such that $x$ and $y$ have equal parity. We shall consider the equivalent equation $$x(x^2{y}^2-2)=c^2+1.$$ If both $x$ and $y$ are even then $x^2{y}^2-2$ is even, whence the l.h.s. is divisible by 4. Thus $c^2\equiv -1(\mathrm{mod}4)$, but this is impossible since $-1$ is not a quadratic residue modulo 4. Let now both $x$ and $y$ be odd; then $x^2{y}^2$ is a positive odd number. Consider two cases: If $x^2{y}^2=1$ then $x^2=1$ and $y^2=1$. Since $x^2{y}^2-2=-1<0$ whereas the product $x(x^2{y}^2-2)$ equals the positive number $c^2+1$, we must have $x=-1$. The possibilities $x=-1,y=1$ and $x=-1,y=-1$ give $a=0,b=1$ and $a=-1,b=0$, respectively. In both cases $c=0$. If $x^2{y}^2>1$ then $x^2{y}^2-2>0$. Squares of odd numbers give remainder 1 upon division by 4, whence $x^2{y}^2-2\equiv 1\cdot 1-2=-1(\mathrm{mod}4)$. Consequently there exists a prime divisor $p$ of $x^2{y}^2-2$ such that $p\equiv -1(\mathrm{mod}4)$. But then $c^2\equiv -1(\mathrm{mod}p)$, which is impossible since $-1$ is not a quadratic residue modulo $p$.