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Old 2021-05-08, 13:37   #4
Dr Sardonicus
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Feb 2017

2·13·191 Posts

Originally Posted by Charles Kusniec View Post
Now, the challenge is to find at least one other form of proof where you use the sum of two squares theorem.
I forgot to point out that the thing is also trivial when n = 1, y^2 = 0 (mod 1).

What I proved was that your given hypotheses imply that

2*n - 1 divides 4*y^2 + 1 if n > 1

2*|n| + 1 divides 4*y^2 + 1, if n < 0.

The fact that every prime divisor p of 4*y^2 + 1 is congruent to 1 (mod 4) is elementary, and does not require the two-squares theorem. Using Fermat's "little theorem,"

p divides (2*y)^(p-1) - 1. From the above,

p divides (2*y)^4 - 1, but p does not divide (2*y)^2 - 1.

Therefore, 2*y has multiplicative order 4 (mod p), from which it follows that p-1 is divisible by 4.

Last fiddled with by Dr Sardonicus on 2021-05-08 at 13:38 Reason: Omit unnecessary words!
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