$a,b \in \mathbb{N} $, $ a b \mid a^2 + b^2$ We need to prove that is a=b ?
Solution: From $ab\mid a^2 + b^2$ $\rightarrow$ $a^2+b^2=k\cdot ab$
then
$$ a^2+b^2=k\cdot ab \quad /\cdot \frac{1}{a} $$ $$ a^2+b^2=k\cdot ab \quad /\cdot \frac{1}{b} $$
we have
$$a+\frac{b^2}{a}=kb$$ $$\frac{a^2}{b}+b=ka $$
Now, we have that from first equation how $a,kb \in \mathbb{N}$ then $\frac{b^2}{a} \in \mathbb{N}$ then $a\mid b^2$ and finally why from this we have that $a\mid b$ ? That part I can't understand.
On
In general, it is not true that if $a|b^2$ then $a|b$. To understand why, see this thread. The statement is true once you do prove that $a = b$ but you haven't yet proven that prior to utilising the step. Therefore you must use other means, such as Aqua's answer, to show that $a = b$.
Let $d= \gcd(a,b)$ then $a=dx$ and $b=dy$ for some relatively prime $x$ and $y$. So we have $$d^2xy\mid d^2(x^2+y^2) \implies xy\mid x^2+y^2\implies x\mid x^2+y^2 \implies x\mid y^2$$
So $x=1$. The same way we see that $y=1$ so $a=b$.