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Let $f(x)$ and $g(x)$ be strictly increasing linear functions from $\mathbb{R}$ to $\mathbb{R}$ such that $f(x)$ is an integer if and only if $g(x)$ is an integer. Prove that for any real number $x$, $f(x) − g(x)$ is an integer.

I let $f(x)=ax+b$, $g(x)=px+q$, $t=\frac{k-q}{p}$, where $k$ is some integer. So $g(t)=k$, and $f(\frac{k-q+1}{p})-f(\frac{k-q}{p})=\frac{a}{p}=m$ is an integer where $a,p>0$. So $f(x)-g(x)=p(m-1)x+b-q=(m-1)g(x)+b-mq$. I'm not sure how to prove it's an integer for all real $x$ though. Please give some hints, thanks!

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  • $\begingroup$ yeah if g(x) is an integer for a particular x, then so is f(x). Think I'm not sure why this works for all real numbers... $\endgroup$
    – daveconked
    Commented Oct 17, 2021 at 5:19
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    $\begingroup$ My instinct is to note that if $h(x)$ is any strictly increasing linear function, then the hypothesis holds for $f(x)$ and $g(x)$ if and only if it holds for $f(h(x))$ and $g(h(x))$; taking $h(x)=f^{-1}(x)$ means that we may assume that $f(x)=x$. $\endgroup$
    – Greg Martin
    Commented Oct 17, 2021 at 5:20

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Note that the statement is symmetric in $f$ and $g$.

You have shown that $\frac a p$ is an integer. By switching the roles of $f$ and $g$, the same proof shows that $\frac p a$ is an integer.

Can you conclude from here? (Hint: $\frac a p \cdot \frac p a = 1$.)

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  • $\begingroup$ oh! so a is just equal to p and the problem is done. That was really right in front of my face and i didn't see it. Thanks! $\endgroup$
    – daveconked
    Commented Oct 17, 2021 at 5:45

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