We wish to find the set of natural numbers that cannot be expressed as a difference between a square and a prime.

e.g.

$1 = 2^2 - 3$

$2 = 3^2 - 7$

$3 = 4^2 - 13$

and so on.

The smallest such number is $16$. The proof that $16$ cannot be expressed as a difference of a square and a prime:

Let $r^2 - p = 16$ for natural number $r$ and prime $p$

$\implies (r-4)(r+4) = p$

$\implies r-4 = 1$ since the two factors of $p$ are $p$ and $1$

$\implies r = 5$ but then $r + 4 = 9$ which is not a prime.

In general, this is true for all $n^2$ where $2n + 1$ is composite since the same reasoning applies. Therefore $49, 100, 144, 169..$ can be seen to belong to this set.

The question is whether all the numbers which cannot be expressed like this are of this form i.e. $n^2$ where $2n + 1$ is composite. A brute force search shows that this holds true for numbers less than 10000.

yes,it is. If $2n+1$ is composite,$\ n$ must of the form $2ij + i + j,\ i, j \in\mathbb{Z},\ 1 \le i \le j\ $, see [Sieve of Sundaram](en.wikipedia.org/wiki/Sieve_of_Sundaram). Let $n = 2ij + i + j,\ $this question is equivalent to prove :$(n+x)^2 - n^2 $ is composite,where $x \in\mathbb{Z^+}$, this is a elementary question, I dont understand this is related to some conjecture. – Mike May 24 '14 at 6:01