prove that the equation $4x^2 + y^2 = 4f^2 + 1$ has a unique solution y = 1 and x = $\sqrt{f} $ if $4f^2 + 1$ is prime. I have no ideas, maybe it's impossible or the statement is false. I found proof thah equation $x^2 + y^2 = p$ has at least 1 solution, but nothing about unique
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1For which range of primes have you verified this numerically (= by brute force)? Saying that you have no ideas rings a bit hollow. – Jyrki Lahtonen Oct 18 '23 at 09:25
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true for n up to 10000 – morz1k3 Oct 18 '23 at 09:34
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1Why hide that piece of information into a comment as opposed to in the question body where everybody can see it? Also, is the solution to be $x=f$ or $x=\sqrt f$ (when the equation should be $x^2+y^2=4f+1$ rather than $x^2+y^2=4f^2+1$)? – Jyrki Lahtonen Oct 18 '23 at 09:57
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It was a duplicate. – Anne Bauval Oct 18 '23 at 15:21
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Any prime p equal to 1 modulo 4 has a unique presentation p = a^2 + b^2, up to the order of the summands. See here: https://proofwiki.org/wiki/Fermat%27s_Two_Squares_Theorem
Dimitri Zvonkine
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as far as I understand this theorem implites that the solution is unique? In the other interpretations of this theoreme they mention only about the existence of solutions, not about unique – morz1k3 Oct 18 '23 at 10:12
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Thanks! I havent even heard about Fermat's Two Squares Theorem/Uniqueness Lemma – morz1k3 Oct 18 '23 at 10:49
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It was a duplicate. In such a predictable case, better look for an older post and link to it, than edit one more answer. – Anne Bauval Oct 18 '23 at 15:21