Show that there is no right triangle whose legs are rational numbers and whose hypotenuse is $\sqrt{2022}$. Show that there is no right triangle whose legs are rational numbers and whose hypotenuse is $\sqrt{2022}$.
My tries:

*

*I used Pythagoras' Theorem to get: $$\sqrt{2022}^2=a^2+b^2 \implies a^2+b^2 = 2022$$
where $a$ and $b$ are the legs of the triangle.
I don't know what to do next: Is there another formula I could use?
I know that $a+b>\sqrt{2022}$ but I don't think this is going to help us much.

hope one of you can help me! thank you!

Quickly ruling out $\ a\ $ being even, we try $\ a\ $ and $\ b\ $ both being odd:
$$ (2k_1+1)^2 + (2k_2+1)^2 = 2022\quad k_1,k_2\in\mathbb{Z}$$
$$ \implies 4({k_1}^2 + {k_2}^2 + k_1 + k_2) + 2 = 2022 $$
$$  \implies {k_1}^2 + k_1 + {k_2}^2 + k_2 = 505, $$
which is impossible, since $\ {k_i}^2 + k_i\ $ is even for $\ i=1,2.$
