How to find integers $a$, $b$ so that $\sqrt{2a^2} = b$

Question

How can I find some integers $a$, $b$ so that $\sqrt{2a^2} = b$?

I want this so I can make a right angle triangle with two sides the same, and all sides have integer length. I tried the integers up to $10$ but none of them have $2\cdot\text{integer}^2$ that I recognise as a square number.

Cheers.

Answer 1

We have $$\sqrt{2a^2} = b\implies a\sqrt2 = b\implies \sqrt 2 = \frac ab$$ which is impossible since $\sqrt2$ is irrational. Hence with the exception of $a=b=0$, no such integers exist.

Answer 2

This question is similar to the proof that $\sqrt2$ is irrational. $\sqrt{2a^2}=b$ gives $\sqrt2a=b$. Then $2a^2=b$, which is not possible when $a$ and $b$ are integers. Of course $a=0, b=0$ would work here, but we probably assuming that $a\ne b$.

Answer 3

Since there is no rational number whose square is $2$ there is no square that's twice a square. But you can get as close as you like. $$ 2 \times 5^2 = 50 = 7^2 + 1, $$ $$ 2 \times 12^2 = 288 = 19^2 -1, $$ $$ 2 \times 29^2 = 41^2 +1, $$ and so on.

See https://en.wikipedia.org/wiki/Square_root_of_2