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$F(m)=(2m^3+2m)/(m^2+1)$ and $g(m)=(m^4+1)/(m^2+1)$ What are the values of $m$ other than $1$ for which solution of both function will be integers. Please tell if there is any formula to find so or any technique?

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. Is $m$ an integer? $(2m^3+2m)/(m^2+1)=2m$, and if $m^2+1|m^4+1$ and $m^2+1|m^4-1$ then $m^2+1|2$ $\endgroup$ – J. W. Tanner Jun 2 at 10:55
  • $\begingroup$ m could be any real number $\endgroup$ – Abhishek Jun 2 at 11:27
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Whenever $m^2+1 \ne 0$, you find that $F(m)=2m$ and $g(m)=m^2-1+2/(m^2+1)$, so $F(m)$ will be an integer for all integers $m$.

Assumig that only integers are allowed for $m$, $g(m)$ will be an integer just for $m \in \{-1,0,1\}$.

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  • $\begingroup$ But $m$ is an integer because if $2m = k \in \mathbb{Z}$, then $g(m)= \frac{k^4+16}{4(k^2+4)}$, hence $k$ is even and $m$ is an integer. $\endgroup$ – Gribouillis Jun 2 at 11:15
  • $\begingroup$ All right, so we don't have to "assume".Thanks! :-) $\endgroup$ – Wolfgang Kais Jun 2 at 11:26
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Well, notice that $\forall x\in\mathbb{Z}$:

$$\frac{2x^3+2x}{x^2+1}=2x\tag1$$

So:

$$2x\in\mathbb{Z}\tag2$$

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