Similar to 223, we could try finding the divisors for \(a^2+1\). However, since \(N\) is larger and \(a^2+1\) doesn’t factor cleanly like 223’s \(a^2-1\), we can bypass factorization entirely by utilizing Barning’s matrices.
These are three specific matrices that generate the “Tree of primitive Pythagorean triples” from a single root. For instance, the root for \(a^2+b^2-c^2=0\) is \((3, 4, 5)\). For barely obtuse triangles, the equation is equals \(-1\) instead, giving us a starting root of \((2, 2, 3)\).
Here is the nice thing: the matrices do not care what the equation equals; they only preserve the left side. Therefore, we simply start at \((2, 2, 3)\), generate new triangles using the three matrices, and stop when the perimeter exceeds the limit. Remember to normalize sides and track what we have seen.
Last question: Can we do this to 223? Theoretical, yes; practical, no. Problem 223 does not have a single node as the root of the tree. If you plug in \(a=1\), the equation becomes \(1+b^2=c^2+1\), meaning \(b=c\). This creates an infinite family of valid starting points \((1, b, b)\). Because of this, it forms a massive, disconnected forest, making a simple graph traversal impossible.
Though, if you really want the \(a^2+1\) factorization approach, note that for some prime \(p\), \({a^2\equiv -1 \pmod {p}}\) gives us \(a^4\equiv 1\pmod p\). That is, all the prime factor of \(a^2+1\) must be either 2 or in the \(4k+1\) format.