# theorem on sums of two squares by Fermat

Suppose that an odd prime number $p$ can be written as the sum

 $p\;=\;a^{2}\!+\!b^{2}$

where $a$ and $b$ are integers.  Then they have to be coprime  .  We will show that $p$ is of the form $4n\!+\!1$.

 $bb_{1}\;\equiv\;1\pmod{p}$

has a solution $b_{1}$, whence

 $0\;\equiv\;pb_{1}^{2}\;=\;(ab_{1})^{2}\!+\!(bb_{1})^{2}\;\equiv\;(ab_{1})^{2}% \!+\!1\pmod{p},$

and thus

 $(ab_{1})^{2}\;\equiv\;-1\pmod{p}.$

Consequently, the Legendre symbol  $\left(\frac{-1}{p}\right)$ is $+1$, i.e.

 $(-1)^{\frac{p-1}{2}}\;=\;1.$

Therefore, we must have

 $\displaystyle p\;=\;4n\!+\!1$ (1)

where $n$ is a positive integer.

Theorem (Thue’s lemma (http://planetmath.org/ThuesLemma)).  An odd prime $p$ is uniquely expressible as sum of two squares of integers if and only if it satisfies (1) with an integer value of $n$.

The theorem implies easily the

Corollary.  If all odd prime factors of a positive integer are congruent to 1 modulo 4 then the integer is a sum of two squares. (Cf. the proof of the parent article and the article “prime factors  of Pythagorean hypotenuses (http://planetmath.org/primefactorsofpythagoreanhypotenuses)”.)

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