Sum of $2$ equal squares also a square 
Is there an integer solution to $a^2 + a^2 = b^2$? Because there's this universift that has this logo of the pytagorean theorem where the two squares are equal, but I don't think it's possible.
 A: We have $$2a^2=b^2,$$ which rearranges to $$2=\frac{b^2}{a^2}=\left(\frac{b}{a}\right)^2.$$
Hence there are integers $a,b$ solving this exactly if $\sqrt{2}$ is rational, which it isn't (see many proofs on wikipedia).  Thus there are no $a,b$ as desired.
A: $2a^2$ is never a perfect square if $a$ is integer. This is because the exponent of $2$ in its prime factorization is odd.
A: If $a, b$ satisfy
$$a^2 + a^2 = b^2,$$
then (for $a, b > 0$) rearranging gives
$$\frac{b}{a} = \sqrt{2}.$$
However, $\sqrt{2}$ is irrational, so there are no positive integers $a, b$ for which this (and hence the original equality) holds, and changing signs shows there are no nonzero integers (regardless of sign) for which it holds.
(Of course, $a = b = 0$ is an integer solution, but surely one you mean to exclude.)
A: OK. So an isosceles right triangle doesn't have integer sides. How about a right triangle with sides that are almost equal?
A nice, and very old, theorem states that a right triangle has rational sides if and only if there is a rational number $\alpha$ such that the sides are in the proportion $\alpha - \dfrac{1}{\alpha} : 2 : \alpha + \dfrac{1}{\alpha}$.
If $\alpha - \dfrac{1}{\alpha} = 2$, then $\alpha = 1 + \sqrt 2$. Wolfram alpha says that the continued fraction representation of $1 + \sqrt 2$ is $[2;\bar 2]$
So we can arbitrarily choose
$$\alpha = [2;2,2,2,2,2,2,2,2,2] = \dfrac{5741}{2378}.$$
This gives us a right triangle with sides of proportion
$$\dfrac{27304197}{13652098} : 2 : \dfrac{38613965}{13652098}$$
$$ 27304197 : 27304196 : 38613965 $$
which is a right triangle with integer sides and is pretty darn close to being an isosceleles right triangle.
