Show that each integer of the form $a^2+b^2$ has all the factors of this form, where $(a, b)$ are distinct integers and relatively prime
If $a^2+b^2$ is prime then it is already proved, since every prime is a factor of its self and $1=0^2+1^2$, but I don't know how to prove for the rest of the case. Thanks!
Assuming that you are talking about numbers of the form $n=a^2+b^2$ with $\gcd(a,b)=1$, then every odd prime $p$ that divides $n$ has the property: $$\left(\frac{-1}{p}\right)=1,\tag{1}$$ since $(1)$ is a consequence of $a^2+b^2\equiv 0\pmod{p}$. However, $(1)$ is equivalent to $$ p\equiv 1\pmod{4},\tag{2}$$ since the Legendre symbol $\left(\frac{-1}{p}\right)$ equals $(-1)^{\frac{p-1}{2}}$.
Now it is possible to prove that any prime $\equiv 1\pmod{4}$ is the sum of two coprime squares - see, for instance, this similar question in which I proved that every prime $p\equiv 1\pmod{3}$ is represented by the quadratic form $a^2+3b^2$ through Fermat's descent - and since the numbers represented by $a^2+b^2$ are a semigroup, due to the Lagrange's identity: $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2\tag{3},$$ we have that every divisor of $n$ is represented by the quadratic form $a^2+b^2$. Notice that $(3)$ is just equivalent to the multiplicativity of the norm on the ring $\mathbb{Z}[i]$ (the gaussian integers).
