# 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?

• Okay, I have posted an answer. Yowza, that was a fun question - thanks for that. :D – Franklin Pezzuti Dyer Dec 1 '18 at 23:42

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 $$c\in\mathbb N$$.