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