Rearrange the given equation: \[a^2+b^2=c^2+1 \Longleftrightarrow (a + 1)(a - 1)= (c + b)(c - b).\] Let \(u=c+b\) and \(v=c-b\), Then \(uv=a^2-1\). Solving for \(b\) yields: \[u-v=(c+b)-(c-b)=2b \Longrightarrow b = \frac{u-v}{2}.\] For \(b\) to be an integer, \(u\) and \(v\) must have the same parity.
Given the perimeter limit \(N\), our bounds become:
- Perimeter constraint: \(a + b + c \leq N \Rightarrow u \leq N - a\)
- Side constraint: \(a \leq b \Rightarrow 2a+v\leq u\).
Finally, iterate \(a\) up to \(N/3\). For each \(a\), generate the divisor pairs \((u,v)\) of \(a^2-1\) utilizing the combined prime factorizations of \(a-1\) and \(a+1\), counting the pairs that satisfy the constraints.