6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Okay, I have posted an answer. Yowza, that was a fun question - thanks for that. :D $\endgroup$ – Franklin Pezzuti Dyer Dec 1 '18 at 23:42
3
$\begingroup$

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 $c\in\mathbb N$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.