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I had a question:

Find all values of $k\in \mathbb{N}‎\cup‎ \{0\}$ such that: $$\frac{k^2-2k}{\sqrt{4k-3}} \in \mathbb{N}‎\cup‎ \{0\}$$

I wanted to solve this problem and I changed this question to a simple one:

Find all values of $m\in \mathbb{N}‎\cup‎ \{0\}$ such that: $$\frac{m^4-1}{2m+1} \in \mathbb{N}‎\cup‎ \{0\}$$

How can I Solve this?

If I can find the solution of second question, then I put $k=m^2+m+1$ for the solution of the first question.

2 Answers 2

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HINT$$\frac{m^4-1}{2m+1} \in \mathbb{Z} \implies \frac{16m^4-1-15}{2m+1}=\frac{(2m-1)(2m+1)(4m^2+1)-15}{2m+1} \in \mathbb {Z}$$ So $$\frac{15}{2m+1} \in \mathbb{Z}$$

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Hint $\ {\rm mod}\ \color{#c00}{2m}\!+\!1\!:\,\ 0 \equiv 16(m^4\!-1)\equiv \overbrace{(2m)^4\!-16}^{\large \color{#c00}{2m}\ \equiv\ -1}\equiv -15,\,$ so $\ 2m\!+\!1\mid 15$