Question

If p and q are positive integers, f is a function defined for positive numbers and attains only positive values such that $f\left(xf\left(y\right)\right)=x^p{y}^q$, then prove that $q=p^2$.,
if p=2 find the value of q

Answer 1

For $x=\frac{1}{f\left(y\right)}$, we have
$f(x.\frac{1}{x})=\frac{1}{f{\left(y\right)}^p}.y^q$ $\Rightarrow$ $f\left(1\right)={\frac{y^q}{\left\{f\left(y\right)\right\}}}^p$
$\Rightarrow$ $f\left(y\right)={\frac{y^{q/p}}{\left\{f\left(1\right)\right\}}}^{1/p}$
For y=1, we have f(1)=1
$\therefore$ $f\left(y\right)=y^{q/p}$ or $f\left(x\right)=x^{q/p}$ (i)
Hence, $f(x.y^{q/p})=x^p.y^q$
Let $y^{q/p}=z\Rightarrow y=z^{p/q}$
$\Rightarrow$ $f(x,z)=x^p{z}^p$
or $f\left(x\right)=x^p$ (ii)
From Eqs. (i) and (ii), we have $x^{q/p}=x^p$
$\Rightarrow$ $\frac{q}{p}=p$ or $q=p^2$

Given, $p=2$

$\therefore q=p^2=2^2=4$