Belarus2022

$$c^2+\tau (c^2)-2=b^2\ge (c+1{)}^2=c^2+2c+1,$$ which means that $\tau (c^2)>2c$. All divisors of $c^2$, except $c$, split into pairs of the form $(d,c^2/d)$, where $d<c$ and $c^2/d>c$, so $\tau (c^2)\le 1+2(c-1)=2c-1$, which contradicts the previously obtained inequality $\tau (c^2)>2c$. Note also that for $a=1$ the number $b$ is equal to zero, which is impossible. Thus $a$ is not a perfect square. Then all positive integer divisors $a$ can be divided into pairs of the form $(d,a/d)$, and hence $\tau (a)$ is an even number. From the equality $a+\tau (a)=b^2+2$ we conclude that the numbers $a$ and $b$ have the same parity, so $a+b$ is even.