# Value of m for which the function will give integers as an output. [closed]

$$F(m)=(2m^3+2m)/(m^2+1)$$ 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?

• 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 $m^2+1|2$ – J. W. Tanner Jun 2 at 10:55
• m could be any real number – Abhishek Jun 2 at 11:27

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 $$m \in \{-1,0,1\}$$.

• 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. – Gribouillis Jun 2 at 11:15
• All right, so we don't have to "assume".Thanks! :-) – Wolfgang Kais Jun 2 at 11:26

Well, notice that $$\forall x\in\mathbb{Z}$$:

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

So:

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