Gaussian integers are the set: $$\mathbb{Z}[\imath] =\{a+b\imath : a,b \in\mathbb{Z} \}$$ With norm: $$\mathrm{N}(a+b\imath)=a^{2}+b^{2}.$$ It satisfies $\mathrm{N}(\alpha\cdot \beta )=\mathrm{N}(\alpha)\cdot\mathrm{N}( \beta )$
The units of $\mathbb{Z}[\imath]$ are precisely: $1,-1,\imath,-\imath$
Result: If $p=4k+1$ ( p : prime), there are $x<p$ such that $p\mid(x^{2}+1)$
$Theorem$: If $p$ is a prime with $p=4k+1$, then $p$ is an sum of two squares. Proof. If $p=4k+1$ there are $x<p$ such that $p\mid (x^{2}+1)$, then $p\mid(x+\imath)(x-\imath)$.
Note that, if $p\mid (x+\imath)$, there are $(a+b\imath)$ such that:
$$(x+\imath)=p(a+b\imath)=pa+pb\imath.$$
This implies that $x=pa$, this is imposible as $x\lt p$.
Therefore $p\nmid(x+\imath)$, also $p\nmid(x-\imath)$.
Then $p$ is not an Gaussian prime, so $p = \alpha \beta$ with $\mathrm{N}(\alpha)\gt 1$ and $\mathrm{N}(\beta)\gt 1$.
If $ \alpha=a+b\imath $ and $\beta=c+d\imath$, we get: $$\mathrm{N}(p)=\mathrm{N}(\alpha \beta)=\mathrm{N}\alpha\cdot\mathrm{N}\beta,$$ which implies: $$p^{2}=(a^{2}+b^{2})(c^{2}+d^{2}).$$
Note that the last equation is an integer, so $(a^{2}+b^{2})|p^{2}$.
By this last equation $(a^{2}+b^{2}),(c^{2}+d^{2})\neq 1,$ so therefore: $$p=a^{2}+b^{2}$$
**
It's beautiful! Does anyone have other proofs ?
**
