Two numbers that cannot both be squares I was wondering where to start with the following question:

Show for $a,b \in \mathbb{N}$ that $a+b^2$ and $a^2+b$ cannot be both squares.

Here $\mathbb{N}$ is the positive integers ($0$ not included).

Consider $a < b$:
Clearly, $b^2 < a+b^2$. Further, we see that
$$
a+b^2 < b + b^2 = b(b+1) < (b+1)^2 
$$
Hence, $b^2 < a+b^2 < (b+1)^2$. Thus, $a+b^2$ is not a square. On the other hand, $a^2+b$ may be a square, depending on the choice of $a$ and $b$. 
If we have $a=b$, then neither $a^2+b$ nor $a+b^2$ are squares, as they would both simplify to $a(a+1)$. 
