proof-explanationpythagorean triplesrational numberstriangles
Show that there is no right triangle whose legs are rational numbers and whose hypotenuse is $\sqrt{2022}$.
My tries:
hope one of you can help me! thank you!
The altitude to the hypotenuse of a right triangle divides the triangle into two similar copies of itself. So knowing that the altitude divides the right angle in the ratio $1:2$, we conclude that the right triangle's acute angles are $30^\circ$ and $60^\circ$.
Let $x$ and $x\sqrt3$ represent the lengths of the legs of the right triangle. We know that the area of the triangle is $\frac12\cdot x\cdot x\sqrt 3=2\sqrt3$, so $\frac12 x^2=2$ and thus $x=2$. The hypotenuse of the triangle is therefore $4$ and the height to the hypotenuse is given by $\frac12\cdot 4\cdot h=2\sqrt 3$ or $h=\sqrt3$.
It can be easily proved by geometry
Consider a right $\Delta ABC$ having hypotenuse $AC$ of constant length which is inscribed in a semi-circle with center $O$ & radius $OA=OE=OC(=AC/2)$ (as shown in figure below).
The right angled vertex $B$ can lie anywhere on the semi-circular arc AC such that $$BD\le EO$$ $$BD\le \color{blue}{\frac{AC}{2}}$$
Best Answer
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 $\ i=1,2.$