F.E. (You run out of these names fast)

For discussing Olympiad Level Algebra (and Inequality) problems
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Zawadx
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Joined:Fri Dec 28, 2012 8:35 pm
F.E. (You run out of these names fast)

Unread post by Zawadx » Wed Feb 15, 2017 9:36 pm

Find all functions $ f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ such that,
$$ f(x+1) = f(x) + 1 $$
and,
$$ f(x^2)=f(x)^2 $$

rah4927
Posts:110
Joined:Sat Feb 07, 2015 9:47 pm

Re: F.E. (You run out of these names fast)

Unread post by rah4927 » Wed Feb 15, 2017 10:48 pm

A brief sketch of the solution.
Plugging in $x=1$ in the second equality gives $f(1)=1$. By induction, it follows that $f(x)=x$ for all positive integers $x$.

Note that $f(x+k)=f(x)+k$ for positive integer $k$. So, $f(m+\frac{k}{m})^2 = (f(\frac{k}{m})+m)^2$. By the second F.E condition, $f(m+\frac{k}{m})^2 = f(m^2+2k+\frac{k^2}{m^2})=m^2+2k+f(\frac{k^2}{m^2})$. Equating the two gives us $f(\frac{k}{m})=\frac{k}{m}$.

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