If p and q are positive integers, f is a function defined for positive number and attain only positive value such that `f(xf(y)) = x^p y^q`, prove that `q = p^2`

Explanation

Given `f(xf(y))=x^(p)y^(q)`
`or x=({f(xf(y))}^(1//p))/(y^(q//p)) " (1)" `
Let `xf(y)=1 or x=(1)/(f(y))`. Then from (1),
`f(y)=(y^(q//p))/({f(1)}^(1//p))`
` or f(1)=(1)/({f(1)}^(1//p))`
` :. f(1)=1`
` :. f(y)=y^(q//p)" (2)" `
Now, `f(xy^(q//p))=x^(p)y^(q). " Put "y^(q//P)=z.` Then
`f(xz)=(xz)^(p)`
`or f(x)=x^(p) " (3)" `
From (2) and (3), `x^(P)=x^(q//p) or p^(2)=q.`