E241 -- Demonstratio theorematis Fermatiani omnem numerum primum formae \(4n+1\) esse summam duorum quadratorum

(Proof of a theorem of Fermat that every number of the form \(4n+1\) can be given as the sum of two squares)


Euler probably already knew the "descent" step, as described in Cox. Here, he does the "reciprocity" step by using the Euler-Fermat theorem to say that \(a^{4n}-b^{4n}\) is divisible by \(4n+1\). Then he factors it as \((a^{2n}+b^{2n})(a^{2n}-b^{2n})\), and says that the second factor cannot be divisible by \(4n+1\); hence the first factor is divisible by \(4n+1\). Beginning in section 12, there seem to be some of those "differences" mentioned by Weil. As indexed by Eneström, this paper concludes on p. 13 of the Novi Commentarii, but Euler's work continues immediately into E242.

According to C. G. J. Jacobi, a treatise with the title: "De numeris qui sunt aggregata duorum quadratorum" was read to the Berlin Academy on October 15, 1750.

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