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This is kind of a reverse question.

A few years back I was presented with a functional equation problem, I don't remember it completely, and now I would appreciate the help of the math.SE hivemind to recreate it.

It concerned a function $f:\Bbb Q^+\to \Bbb Q$, with $\Bbb Q^+$ being the strictly positive rationals. The problem gave two identities for $f$, one of them being $f(x) = f(x^{-1})$. I can't for the life of me recall what the second one was. However, I believe it related $f(x)$ and $f(x+1)$. There is a possibility that $f(1) = 1$ was also included as a restraint.

What I do remember, though, is the solution. With $a, b\in \Bbb Z$ coprime, $f(\frac{a}{b}) = ab$. Also, that the two identities together is in some way related to the Euclidian algorithm.

Long story short, does anyone know what the second identity is / could be?

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3 Answers 3

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This has nothing to do with $f(x+1)$, but what about: $$f(x^2)=f(x)^2$$

Edit: what about this one: $$f(x+1)+f(x-1)=2\cdot f(x)$$

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  • $\begingroup$ $f(x)=0$ and $f(x)=1$ are each solutions for both your suggestions as well as the original $f(x) = f(x^{-1})$ $\endgroup$
    – Henry
    Commented Oct 12, 2012 at 19:04
  • $\begingroup$ @Henry if they are identities, they should work for all $x=\frac{a}{b}$ where $a$ and $b$ are coprime. Or am I missing something? $\endgroup$
    – Alex
    Commented Oct 12, 2012 at 19:38
  • $\begingroup$ My point was that although $f(\frac{a}{b}) = ab$ is a solution to your expressions (which I upvoted), it is not the only one $\endgroup$
    – Henry
    Commented Oct 12, 2012 at 19:46
  • $\begingroup$ @Henry I see what you mean. But how is $f(x)$ a solution for the original identity? (And thanks for the upvote :)) $\endgroup$
    – Alex
    Commented Oct 12, 2012 at 19:50
  • $\begingroup$ If $f(x)$ is constant then $f(x) = f(x^{-1})$ $\endgroup$
    – Henry
    Commented Oct 12, 2012 at 19:56
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The 2nd condition can be:

$x \cdot f(x+1) = (x+1) \cdot f(x)$

This gives, $f(m) = m$ for all $m \in \mathbf{Z}$ when used with the fact $f(1) = 1$.

To extend this to positive rationals, we can use $f(x) = f(x^{-1})$ and a repeated use of Euclid's division algorithm (as you mentioned).

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  • $\begingroup$ This looks like the right one to me. Thanks a lot. $\endgroup$
    – Arthur
    Commented Oct 13, 2012 at 9:08
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If $a$ and $b$ are coprime then $a+b$ and $b$ are also coprime, so you get $$ f\left(\frac{a}{b} + 1\right) = f\left(\frac{a+b}{b}\right) = (a+b)b = ab+b^2 \text{ if } a,b \text{ coprime} $$ and $$ f\left(\frac{b}{a} + 1\right) = f\left(\frac{b+a}{a}\right) = (b+a)a = ab+a^2 \text{ if } a,b \text{ coprime} $$ and thus generally $$ \frac{f(q^{-1} + 1) - f(q^{-1})}{f(q+1) - f(q)} = q^2 $$

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