## Task 4 - 100834

Let $a, b$ be natural numbers, and let $a > b$.

Give conditions for $a$ and $b$ such that the following holds: The difference of the squares of $a$ and $b$ is a prime number if and only if all these conditions are satisfied!
Due to $a>b$, it holds that $a^{2}-b^{2}>0$. Due to $(a-b) \geq(a+b)$, $a^{2}-b^{2}=(a-b)(a+b)$ is a prime number if and only if $a-b=1$ and $a+b$ is a prime number.