Another Reworded APMO Problem!

We consider 2 cases:

Case 1:

If $a$ or $b$ is $0$ , then the other would be a perfect square. Which is obviously a solution.

Thus, a solution is $(a, b)=(k^{2}, 0)=(0, k^{2})$ for some integer $k$ .

Case 2:

Let $a$ and $b$ both be positive integers.

If both of them are positive, then $a^{2}+4b=(a+2n)^{2}$ for some positive number $n$ . So $b=na+n^{2}>a$ . But similarly, $a>b$ , a contradiction. Thus there are no solutions for this case.

For case 1, there are $31 \times 2 +1=63$ possible solutions.

For case 2, there are $0$ possible solutions. And

$63+0=\boxed{63}$

Feel free to ask if there are any questions.

First,since the question asks for non-negative pairs $(a,b)$ , we are motivated to put $a=0$ and likewise $b=0$ .It is easy to see that there are $63$ solutions with one of them equal to $0$ .

Now,we restrict the values of $a,b$ to positive integers only.A common approach with this type of problem is to use easy inequalities.Note that $a^2$ is a perfect square.So,if $a^2+4b$ is a perfect square,it must be greater than $a^2$ . The next closest perfect square to $a^2$ is $(a+1)^2$ . So, $a^2+4b\ge (a+1)^2$ i.e. $4b\ge 2a+1$ Now similarly,we can get $4a\ge 2b+1$ Now since we can do nothing better,we add the inequalities with the hope of a contradiction.But alas,we get $a+b\ge 1$ which does not help us in any way.Nevertheless,the method looked promising,so let's strengthen the inequality a bit.We look at the next closest perfect square to $a^2$ , which is $(a+2)^2$ .So if $a^2+4b\ge (a+2)^2$ we get $4b\ge 4a+4$ and similarly $4a\ge 4b+4$ This time,adding the inequalities gives us $0\ge 8$ and we know that's not true.

But what if $a^2+4b$ is smaller than $(a+2)^2$ ?Well,if it is smaller than $(a+2)^2$ ,it can only be equal to $(a+1)^2$ . Now a little case checking will show that there are no more $(a,b)$ and this part of the proof is left to the reader.