We will show that all divisors of x2 +1 have the form 4m + 1 or 2(4m + 1).
We have already shown in the chapter about sequences of Farey that if
n divides x2 +1 then n can be represented as a sum of two integer squares, without common divisor.
In other words, if n divides x2 +1, then n = a2 + b2 (for certain a and b).
a and b are not both even, for a and b do not have a common divisor.
Thus a and b are both odd or a is odd (even) and b is even (odd).
So n = a2 + b2 = (2k+1)2 + (2m+1)2 = 4(k(k+1) + m(m+1)) + 2 = 8(k(k+1)/2 + m(m+1)/2) + 2 or
n = a2 + b2 = (2k+1)2 + (2m)2 = 4(k(k+1) + m) + 1. (for certain integers m and k).
This accomplishes the proof.