Yes: let $a=10^{100}+1002, b=1001$. In order for $a,b,c$ to be the sides of a triangle, we need $a-b<c<a+b$. But as $10^{100}$ is a square, there are no squares between $10^{100}+1$ and $10^{100}+2\cdot 10^{50}+1$. The first number is $a-b$ and the second number exceeds $a+b$, so $c$ cannot be a square.