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 $4n+1$.

Since  $p\nmid b$,  the congruence

$$b{b}_1\equiv \mathrm{\hspace{0.33em}1}(modp)$$

has a solution $b_1$, whence

$$0\equiv p{b}_1^2={(a{b}_1)}^2+{(b{b}_1)}^2\equiv {(a{b}_1)}^2+1(modp),$$

and thus

$${(a{b}_1)}^2\equiv -1(modp).$$

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

$${(-1)}^{\frac{p-1}{2}}=\mathrm{\hspace{0.33em}1}.$$

Therefore, we must have

$p=\mathrm{\hspace{0.33em}4}n+1$(1)

where $n$ is a positive integer.

Euler has first proved the following theorem presented byFermat and containing also the converse of the above claim.

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

The theorem implies easily the

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