Take $n \in \mathbb{N}$. Now it is obviously, that every square $(n+1)^2$ with is the sum of a square, i.e. $n^2$ and an odd number, i.e. $2n+1$. Since every odd number $2n+1$ is either a prime number or not a prime number (and, of course, there exist infinitely many odd primes and infinitely many odd composite numbers), there are infinitely many odd prime numbers as well as infinitely many odd composite numbers, so there are infinitely many squares $(n+1)^2$ which are a sum of a square and a prime number as well as a sum of a square and an odd composite number, which is not prime. This solves a. The solution for b also needs, that for all $a$ with $a<n$, $(n+1)^2-a^2=(n+a+1)(n-a+1)$ is an odd composite number and hence, $(n+1)^2$ is not a sum of an ordinary (is it okay like this, Xorg? ) square and a prime number.

Hallo, for b it only suffices to prove that (n +1)^2 - n^2 is composite for infinitely many n, because the difference between 2 squares is surely composite by (a-b) (a+b) ;except for b equals a-1 . For n = 2^(2a+1) - 1, we have 2n + 1 = 2^(2a+2) - 1 which factors for infinitely many a and is therefore not a prime .