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Consider the following right-angled triangle, in which the angle between the side $\text{AC}$ and $\text{CB}$ is a right angle. Let $a,b,c$ be the length of the sides $\text{BC, AC, AB}$ respectively. Assume $a,b,c$ are integers.

Which of the following is correct?

- At least one of $a,b$ must be of even length.
- At least one of $a,b$ must be of odd length.
- $c$ must be of even length.
- It is possible that the length of every side is odd.

3 votes

We have proven the following proposition in the homework $2.$

Suppose $a,b,c\in \mathbb{Z}.$ If $a^2+b^2=c^2,$ then $a$ or $b$ is even.

An interesting consequence of this result is that in any right triangle in which all sides have integer length at least one of the two shorter sides must be of even length.

Suppose $a,b,c\in \mathbb{Z}. $ If $a^2+b^2=c^2,$ then $a$ or $b$ is even.

This result can be re-phrased as following :

$\textbf{Cool Application I:}$ Sums of odd perfect squares. Can a sum of two perfect squares be another perfect square? Sure; for example, $3^2+4^2=5^2, \; 5^2+12^2=13^2 \; 6^2+8^2=10^2, \; 7^2+24^2=25^2.$

However, no matter how much you try, you won't find any examples in which the two perfect squares on the left are both odd. Your task is to prove this, i.e.,

*Prove that a sum of two odd perfect squares is never a perfect square.*

(An interesting consequence of this result is that in any right triangle in which all sides have integer length at least one of the two shorter sides must be of even length.)