Find all non-negative integers $a, b$ satisfying $|4a^2 - b^{b+1}| \leq 3$ Find all non-negative integers $a, b$ satisfying $|4a^2 - b^{b+1}| \leq 3$.  
I have been trying a simpler case $4a^2 - b^{b+1} = 0$. I found that when $b$ is odd then $b^{b+1}$ have to be in form $b^{b+1} = c^2$ where $c$ is integer. So we have $(2a + c)(2a - c)=0 \implies 2a = c$ (otherwise $a$ would be negative). However, $c$ is odd because $b$ is also odd (odd number times odd number is odd). But this means that $a$ (since $2a = c$) isn't integer and that is contradiction. Therefore in this simpler case $b$ can't be odd. 
But above applies only for specific simpler case. For example for $a = b = 1$ original inequality holds.
Can someone help me with this please?
 A: For the sake of simplicity, I will consider positive integers $a,b$.
First, let's consider the case of $4a^2-b^{b+1}=0$. You've already noticed that $b$ cannot be odd, and so we shall consider only the case of even $b$. If $b$ is even, then $b+1$ is odd and so $b^{b+1}$ is a perfect square iff $b$ is a perfect square, so we have that $b$ is a perfect square. Since it is an even perfect square, we may write $b=4c^2$ and $b^{b+1}=4\cdot 4^{4c^2}\cdot c^{8c^2+2}$ and we may let $a=4^{2c^2}\cdot c^{4c^2+1}$ for a solution. Thus, we have found one possible solution set:
$$(a,b)=(4^{2c^2}\cdot c^{4c^2+1},4c^2)$$

Now suppose that $4a^2-b^{b+1}=1$. Then $(2a+1)(2a-1)=b^{b+1}$, and so $b$ must be odd, meaning that $b+1$ is even and $b^{b+1}$ is a perfect square. But $4a^2=b^{b+1}+1$ is also a perfect square, giving a contradiction. So $4a^2-b^{b+1}\ne 1$.
Similarly, suppose that $4a^2-b^{b+1}=-1$, or $4a^2=b^{b+1}-1$. We have again that $b$ must be odd, and so $b+1$ is even and $b^{b+1}$ is a perfect square, which is a contradiction since $b^{b+1}-1$ is a perfect square.

Now suppose that $4a^2-b^{b+1}=2$, or $4a^2=b^{b+1}+2$. Then $b$ must be even, and we have that the $2$-adic valuation of $b^{b+1}+2$ is equal to $1$. But the $2$-adic valuation of $4a^2$ is at least $2$, giving a contradiction.
Use the same reasoning for the case of $4a^2-b^{b+1}=-2$.

Suppose that $4a^2-b^{b+1}=3$, or $4a^2=b^{b+1}+3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}+3$ is also a perfect square, which is only possible if $b^{b+1}=1$. This gives the solution pair 
$$(a,b)=(1,1)$$
and no others.
Finally, suppose that $4a^2-b^{b+1}=-3$ or $4a^2=b^{b+1}-3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}-3$ is also a perfect square, which can only occur if $b^{b+1}=4$, but this never happens for positive integers $b$.

We are done! We have only the solutions $(1,1)$ and 
$$(4^{2c^2}\cdot c^{4c^2+1},4c^2)$$
for $c\in\mathbb N$.
