Pythagorean Triple where $a=b$?

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 $\Rightarrow\Leftarrow$.

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.

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.