Sum or difference... or both

Since $(n+1)^2 - n^2 \; = \; 2n+1 \hspace{1cm} (n+1)^2 - (n-1)^2 \; = \; 4n \hspace{2cm} n \ge 0$ we see that odd numbers and multiples of $4$ can all be written as a difference of two squares. Moreover, since $a^2 - b^2 = (a+b)(a-b)$ and $a+b,a-b$ have the same parity, we deduce that all differences of two squares are either odd or multiples of $4$ . Thus the only numbers that cannot be written as a difference of two squares are the numbers that are congruent to $2$ modulo $4$ .

Since $a^2 + b^2 = (a+ib)(a-ib)$ , the numbers that can be written as a sum of squares are going to be identified by considering factorisation in the Gaussian integers $\mathbb{Z}[i]$ . A prime integer $p \in \mathbb{N}$ is prime in $\mathbb{Z}[i]$ if and only if $p \equiv 3 \pmod{4}$ ; the prime $2$ and primes $p \equiv 1 \pmod{4}$ can all be written as a sum of squares of two integers. Thus an integer $n$ can be written as a sum of two squares precisely when its prime factorisation in $\mathbb{Z}$ does not contain an odd prime, congruent to $3$ modulo $4$ , which is repeated an odd number of times.

If $n \equiv 6 \pmod{8}$ , then the prime factorisation of $n$ over $\mathbb{Z}$ must contain an odd number of primes congruent to $3$ modulo $4$ , and this means that such numbers cannot be written as a sum of two squares. There are $12$ such numbers between $1$ and $100$ : $6$ , $14$ , $22$ , $30$ , $38$ , $46$ , $54$ , $62$ , $70$ , $78$ , $86$ , $94$ .

If $n \equiv 2 \pmod{8}$ , then all but two of these numbers have prime factorisations which mean that they can be written as a sum of two squares. The two exceptions are $42 = 2\times3\times7$ and $66 = 2\times3\times11$ .

Thus there are $12+2 = 14$ numbers which cannot be written as either a difference or a sum of two squares, and hence there are $\boxed{86}$ numbers that can.