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Prove that for all positive integers $n \geq 17$ such that $\left\lfloor \frac{n}{4} \right\rfloor \equiv 0 \pmod 2$, there exists positive integers $a, b$ and an even positive integers $c$ such that the following are satisfied:

$$ a+b = 4c-1$$ and $$\left\lfloor \frac{n}{4}\right\rfloor + c \equiv 0 \pmod {ab}$$

To me, this seems logical but I've spent the past month trying to prove it but haven't been able to make any progress. I feel really stumped. Any help is appreciated, thanks!

Andrew
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  • How does that seem "logical"? And what is the purpose behind the $\lceil n/4\rceil$ when that expression just runs over all positive integers as well? – Hagen von Eitzen Nov 30 '16 at 09:47
  • This is just a lemma for a larger problem that I am proving so that's where the $\lceil n/4 \rceil$ is from and thank you for your edits, I am still learning LaTex. – Andrew Nov 30 '16 at 09:53
  • Might help to note that $\left\lceil\frac{n}{4}\right\rceil\equiv0\pmod2 \iff n\equiv0,5,6,7\pmod8$. – barak manos Nov 30 '16 at 10:00
  • After your last edits, it seems you could still start with "for all even integers $m\ge 4$ ..." and replace $\lfloor n/4\rfloor $ with $m$ – Hagen von Eitzen Nov 30 '16 at 10:10

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For $n$ as in the problem statement let $c:=\lfloor \frac n4\rfloor$, $a=2c$, $b=2c-1$.

Hagen von Eitzen
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