Proving $p^2=q$ given a functional equation

Question

Question:

If $p,q$ are positive integers, $f$ is a function defined for positive numbers and attains only positive values such that $f(xf(y))=x^py^q$, then prove that $p^2=q$.

My solution:

Put $x=1$. So, $f(f(y))=y^q$, then evidently, $f(y)=y^{\sqrt{q}}...(1)$ satisfies this.
Now, put $x=y=1$, to get $f(1)=1$
Now, put $y=1$. So, $f(x)=x^p...(2)$

Equalising $f(a)$ for an arbitrary constant $a$ from the two equations $(1)$ and $(2)$, we get: $a^{\sqrt{q}}=a^p$ or $p^2=q$. $\blacksquare$

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Is this solution correct? I am particularly worried because I have solved this six marks question in a four line solution which wouldn't make my prof very happy...

Answer 1

There is a major gap in your argument.

You've correctly argued that any $f$ satisfying the functional equation $f(x f(y)) = x^p y^q$ must also satisfy the functional equation $f(f(y)) = y^q$.

It is correct that $f(t) = t^\sqrt{q}$ is one solution to the functional equation $f(f(y)) = y^q$.

However, you have made no attempt to show it is the only solution. In fact, other solutions to this functional equation exist; a simple one is that $f(t) = 17/t$ satisfies $f(f(y)) = y^q$ when $q=1$. More complicated solutions exist too.

Thus, you have not shown that any $f$ satisfying $f(x f(y)) = x^p y^q$ must also satisfy $f(t) = t^\sqrt{q}$. Some possible ways to continue are:

• Find a proof that doesn't rely on this assumption
• Find out what the other cases are and prove that they also have $p^2 = q$
• Prove that functions satisfying $f(x f(y)) = x^p y^q$ must actually be of the form $f(t) = t^\sqrt{q}$.

Answer 2

Your solution is not conclusive, since $f(f(y))=y^q$ has infinitely many solutions. But if you want to make that professor unhappy with a short solution, you can do that: replacing $x$ by $f(x)$ in your equation, you get $$f(f(x)f(y))=f(x)^py^q.$$ The LHS is symmetric in $x,y$, so you must have also $$f(x)^py^q=f(y)^px^q,$$ meaning
$$f(x)^px^{-q}=f(y)^py^{-q}.$$ So that must be a constant, and you arrive at $$f(x)=c\,x^{q/p}.$$ If you plug this into your original equation, you obtain the two conditions $c^{1+q/p}=1$ and $q/p=p$, i.e. $p^2=q$ and $c=1$.