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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

gt6989b
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Patrik Kula
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2 Answers2

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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.

saulspatz
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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 $m=\pm4^{2a^2}a^{4a^2+1}$.

Servaes
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    I was going to update my answer to say it comprised all solutions with $n$ even, but now I don't have to. +1 – saulspatz Oct 17 '18 at 15:39
  • thanks :D the first I could do but in the second part I didnt noticed that I can get the four out of that abs. value, thanks very much :) – Patrik Kula Oct 17 '18 at 16:43