Question involving square equality between fractions and square roots

Question

Find the values of the constants $p$ and $q$ such that $$\frac{\sqrt{p}}{\sqrt{p}+2p} = \frac{2\sqrt{p}-q}{3p+q} \tag{$p,q\ge0$}$$ How would you solve this? I've tried everything...

Answer 1

There are a condition missing since if there's only one equation for 2 unknowns, the number of solution is infinite. I've got $$ (p,q)=(p,\frac{p\sqrt{p}+2p}{2p+2\sqrt{p}}) $$ For example, $(2,1)$ works

EDIT We've got $$ \sqrt{p}(3p+q)=(\sqrt{p}+2p)(2\sqrt{p}-q) $$ which is legit since the divisers are positives $$ 3p\sqrt{p}+q\sqrt{p}=2p-q\sqrt{p}++4p\sqrt{p}-2pq $$ $$ 2q\sqrt{p}+2pq=2p+p\sqrt{p} $$ $$ q=\frac{p\sqrt{p}+2p}{2p+2\sqrt{p}} $$ I assumed that the question was for p,q positives integers, otherwise the number of solution is too big. You can easily see that for p=2, q=1

Answer 2

Let $a = \sqrt{p}$, and $b = q $, then: $\dfrac{a}{a + 2a^2} = \dfrac{2a - b}{3a^2 + b}$.

So: $3a^3 + ab = 2a^2 - ab + 4a^3 - 2a^2b$ or:

$a^3 + (2 - 2b)a^2 - 2ab = 0$. Thus:

$a(a^2 + (2 - 2b)a - 2b) = 0$, since $a \neq 0$, we have: $a^2 + (2 - 2b)a - 2b = 0$, and use quadratic formula:

$\triangle' = (1 - b)^2 + 2b = 1 + b^2$. So:

$a = -1 + b + \sqrt{1 + b^2}$ ( the other root is negative so we can't take it ). So:

$p = a^2 = \left(-1 + q + \sqrt{1 + q^2}\right)^2$.

Specifically, choose $q = 1$, then $p = 2$.