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We have to verify that for any integer n^2 + 1, the odd prime divisors are of the form 4k + 1.
According to Euler's theorem, if p is an arbitrary odd prime, then x^2`-=` -1(mod p) has a solution p`-=` 1 (mod 4)
We'll use this to arrive at the proof.
Let the integer n^2 + 1 be divisible by an odd prime divisor p, n^2 + 1`-=` 0(mod p)
=> n^2 `-=` -1(mod p)
The solution to this is p`-=` 1(mod 4)
or p = 4k + 1
This proves that all odd prime divisors of n^2 + 1 are of the form 4k + 1
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