Pythagorean Triple where $a=b$?

Question

I cannot find even a single webpage mentioning this topic. I'm a programmer and I'm looking for a 45-45-90 triangle where all of the sides are whole numbers. In the video I am watching, they say to use $ a = 10 $, $ a = 10 $, $ c = 14 $ because $ 10 \sqrt{2} $ is close enough to $ 14 $. In my program I am worried this could have serious consequences because it's not accurate.

Does there exist a case where $ 2 a^2 = c^2 $ where a and c are whole numbers?

If it does not exist, why? Does it revolve around the fact that $ \sqrt{2} $ is irrational?

Answer 1

Yes, it does revolve around the fact that $\sqrt{2}$ is irrational.

For if there were integers $a$ and $c$ such that $a^2 + a^2 = c^2$, then $2a^2 = c^2$, or $2 = \left(\frac{c}{a}\right)^2$. Therefore $\sqrt{2}$ is rational $\Rightarrow\Leftarrow$.

Answer 2

There is no isosceles right triangle with integer sides. That's equivalent to the fact that the square root of $2$ is irrational. The sequence $$ 3/2, 7/5 , 17/12, \ldots $$ provides rational approximations to $\sqrt{2}$ as accurate as you wish. Here $a/b$ corresponds to the solution $$ a^2 - 2b^2 = \pm 1 $$ to Pell's equation.

Far enough along on that sequence you should be able to find an approximation good enough for your program.

Answer 3

Yes, it is just the fact that $\sqrt{2}$ is irrational. Suppose there would be such a right-triangle. Then $n=1$ would be a congruent number, i.e., the area of an right-triangle with rational sides. By Fermat, for exponent $4$, it isn't.