If $a$ and $b$ are odd integers, then $\sqrt{a^2+b^2}$ is irrational

Question

If $a,b\in\mathbb{N}$ are odd

then demonstrate: $$ {\sqrt{a^2 + b^2}} \not\in \mathbb{Q}$$

I try to guess that $$ {\sqrt{a^2 + b^2}} \in\mathbb{Q}.$$ Then i write $$ {\sqrt{a^2 + b^2}= m/n}.$$ After that: $$ {n\sqrt{a^2 + b^2}= m}$$ , I raised at squared and i have like $$ n^2(a^2+ b^2)=m^2 $$ and i thought that $$ (a^2 +b^2)$$ is even so m^2 is even. After this I write $$m^2= 4k^2.$$ In the and I have this ecuation $$a^2 + b^2 = 4k^2/ n^2$$ This fraction is irreducible..I think

Answer 1

Hint: $(2m+1)^2+(2n+1)^2=2(2k+1)$. Show that this number cannot be written as $\dfrac{p^2}{q^2}$ with $(p,q)=1$.

Answer 2

Hint: The square of an odd number is of the shape $4k+1$, indeed $8k+1$.

Answer 3

You may remember that Pythagorean triplets of a right triangle are of the form, a²-b², 2ab and a²+b², the latter one being the hypotenuse. Well, the second terms says it all...