diophantine equation $|4m^2-n^{n+1}| \le3$

Question

please can someone give me a hint in this equation?

$|4m^2-n^{n+1}|\le3$ for non zero integers, find for which numbers this equation holds

I found roots as $m,n=0; m,n=1; m=0$ and $n=1$

I tried prove as those are only one with odd or even integers but it was with out success

and I tried substitution $k=n+1$ and prove it with binomial coefficients and it wasn't good

Thanks for hints :D

Answer 1

If $n$ is odd, say $n=2k-1$, then $$|4m^2-n^{n+1}|=|(2m)^2-((2k-1)^k)^2|,$$ which is the difference of an odd and an even square. For the inequality to hold we must have $2k-1=\pm1$ and $-1\leq m\leq 1$, yielding the following solutions $(m,n)$: $$(0,\pm1),\ (\pm1,\pm1),$$ most of which you already found.

If $n$ is even, say $n=2k$, then $$|4m^2-n^{n+1}|=4|m^2-2^{2k-1}k^{2k+1}|,$$ so for the inequality to hold this must be zero, i.e. $2^{2k-1}k^{2k+1}$ must be a square. Equivalently $(2k)^{2k+1}$ must be a square, which is the case if only if $k$ is twice a square, i.e. $k=2a^2$ for some integer $a$. Then $n=4a^2$ and so $$n^{n+1}=(4a^2)^{4a^2+1}=4(4^{2a^2}a^{4a^2+1})^2,$$ which shows that we must have $m=\pm4^{2a^2}a^{4a^2+1}$.

Answer 2

Here are infinitely many solutions, but this is just a start. If $n = 4k^2,$ then $$n^{n+1}=4k^2(4k^2)^{n}=4k^24^nk^{2n}=4k^2(2k)^{2n}$$ so we can take $m=k(2k)^n$ to get $4m^2=n^{n+1}.$

I have no idea how to find all solutions to the inequality.