The circle passing through \((0, 0),(N,0),(0,N),\) and \((N,N)\) has the equation:\[\Bigg(x-\frac{N}{2}\Bigg)^2+\Bigg(y-\frac{N}{2}\Bigg)^2=\frac{N^2}{2}\] Multiplying by \(4\) yields: \[(2x-N)^2+(2y-N)^2=2N^2\] Let \(U=2x-N\) and \(V=2y-N\). The problem reduces to finding the number of integer solutions \((U,V)\) to \(U^2+V^2=2N^2\).
Let the prime factorization of \(N\) be: \[N=2^a\prod p_i^{e_i}\prod q_j^{f_j}\] where \(p_i\equiv 1\pmod{4}\) and \(q_j\equiv 3\pmod{4}\). Consequently, the prime factorization of \(M=2N^2\) is: \[M=2^{2a+1}\prod p_i^{2e_i}\prod q_j^{2f_j}\] By Jacobi’s two-square theorem, the number of integer solutions to \(U^2+V^2=M\) is \(r_2(M)=4\prod (E_i+1)\), where \(E_i\) are the exponents of \(p_i\), provided the exponents of \(q_j\) are even (which \(2f_j\) trivially satisfies). Thus: \[f(N)=4\prod (2e_i+1)=420\Longrightarrow \prod(2e_i+1)=105\] Factoring \(105\) defines the required multisets of exponents \(e_i\) for primes \(p_i\equiv 1\pmod{4}\):
- \(105\Longrightarrow \{52\}\)
- \(3\times 35 \Longrightarrow \{1, 17\}\)
- \(5\times 21 \Longrightarrow \{2, 10\}\)
- \(7\times 15 \Longrightarrow \{3, 7\}\)
- \(3\times 5\times 7 \Longrightarrow \{1, 2, 3\}\)
Therefore, \(N=m\prod p_i^{e_i}\) where \(m\) has no prime factors congruent to \(1\pmod{4}\), and the exponents \(e_i\) strictly match one of the multisets above.