background
Again “probfromthebook”, this time discussing the classes of primes congruent to \(1,3 \mod 4\). Let’s call these classes \(P_1,P_3\). Let’s also define the set \(Q_2\): numbers which are expressible as a sum of squares.
Fermat’s remarkable theorem says:Theorem. \[P_1\subset Q_2\]
Lemma. Let \(p=4k+1\) be prime. \(((2k)!)^ 2 \equiv -1 \mod 4k+1\)
Proof. \(((2k)!)^ 2 \equiv (4k)!\equiv -1 \mod 4k+1\) by “wilson’s theorem”. To prove it from basic facts, note that \(x^{2}\equiv 1\) has exactly \(2\) solutions, namely \(x=\pm 1\), because \(\mathbb{F}_p\) is a field. Thus, all other things are not self-inverses, and we can pair them up with their inverses. This just leaves \(\pm 1\).
Lemma. Let \(a\perp p\). Then there exist \(x,y \le \sqrt{p}\) and \(\sigma \in \pm 1\) such that \[a\equiv \sigma x^{-1}y \mod p\]
Proof. pidgeon
Corollary. If you put together the two lemmas and do some math you get the desired result.
Proof. exercise for reader
Theorem. Let \(p\in P_3\), and let \(x,y\in \mathbb{N}\) with \(x\perp p\). Then \(p\not\mid x^{2}+y^{2}.\)
In particular, \(p\not \mid 1+n^2\) for any \(n\in \mathbb{N}\).
Proof. exercise for reader.
Theorem. \(P_1,P_2\) are infinite
Proof. exercise for reader. Kind of like the proof that there are infinitely many primes.
Theorem. \(Q_2\) is the set of numbers \(n\) with \(\nu_p(n)\in 2\mathbb{Z}\) for all \(p\in P_3\).
Proof. clear corollary of stuff we talked about so far.
actual problem
Theorem. \(4mn-m-n\) is not a perfect square
Proof. Proof by contradiction. Factor as \[(4m-1)(4n-1) = 4k^2 + 1.\] Then, all prime factors of \(4k^2+1\) should be in \(P_1\). However, the LHS is a product of things which are \(3\mod 4\). Contradiction.