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I can reduce it using the fact that square root of a number is rational if it is a square number.
Now it would be something like :
If $ab=k^2$ for some $k$ in $N$ then show that $frac{a}{b}=m^2$ for some $m$ in $N$

Where $a$,$b$ belongs to $N$

$N$ is the set of natural numbers.

I have tried prime factorisation also
But do not know how to proceed.

CLAIM

If $a$,$b$ belongs to $N$ and if $sqrt ab$ is integer then prove that $sqrt frac{max(a,b)}{min(a,b)}$ is also an integer.
Where $N$ is the set of natural numbers.

Hint $ $ Employ $,bsqrt{a/b} = sqrt{ab}.,$ This is a special case of a general method. Let's view it that way.

Observe that $, overbrace{bx^2 = a}^{large x = sqrt{a/b}}!!!!overset{large times b}iff! overbrace{(bx)^2 = ab}^{large, bx = sqrt{ab}} $ for $,bneq 0$

We scaled the LHS by $b$ to get $,X^2 = ab, X := bx,$ which is now monic (lead coeff $=1).,$ This is a special case of the monicization method used in the AC method for factoring non-monic quadratics.

The same method also works for higher-degree polynomials (as I explain in the linked post), which leads to higher-degree generalizations of the above identity for square-roots.

Generally the method shows that any algebraic number can be written as $,alpha = beta/n,$ where $,ninBbb Z,$ and $,beta,$ is an algebraic integer (i.e. a root of a monic $f(x)inBbb Z[x]),$ e.g. above $,sqrt{a/b} = sqrt{ab}/b,,$ where $,sqrt{ab},$ is a root of the monic $,x^2!-ab,$ so is an algebraic integer.

In higher-degree cases such equations are generally less obvious, so it is worth being aware of the general viewpoint (even though it is a bit overkill for this particular case).

This hint will help you not only prove something you want to, but also state more clearly what you meant.

If $a,,b$ are rational with $bne 0$ then $sqrt{frac{a}{b}}=frac{sqrt{ab}}{b}$ is the ratio of two rationals, the denominator non-zero. Any such ratio is rational (proof is an exercise).

Your claim works if and only if $b$ is rational.

Note that

$frac{sqrt{ab}}{sqrt{dfrac{a}{b}}}=b$

If numerator is rational then denominator is rational too.

Counterexample for the case when $b$ is irrational take
$a=2sqrt{2}$ and $b=sqrt{2}$

Here $sqrt{ab}=2$ however $sqrt{dfrac{a}{b}}=sqrt{2}$

First, point out that your statement doesn’t hold for $a,b,k,m in mathbb{N}$.

Take for example $(a,b) = (2,18)$. Then $k = sqrt{ab} = 6 in mathbb{N}$, but $m = sqrt{frac{a}{b}} = frac{1}{3} notin mathbb{N}$.

But the following statement does hold:

If $ab = k^2$ with $a,b,k in mathbb{Q}$, then $frac{a}{b} = m^2$ for some $m in mathbb{Q}$, where $mathbb{Q}$ is the set of rational numbers.

Proof: If $ab = k^2$, then $frac{a}{b} = frac{ab}{b^2} = frac{k^2}{b^2} = (frac{k}{b})^2$, so $m = frac{k}{b}$ is rational, as desired.

An "easy" algebraic proof:

if (assuming that $a$ and $b$ are rational numbers) $sqrt {ab}={frac{x}{y}}$, then $ab$ is a perfect "square fraction" i.e. $ab=frac{x^2}{y^2}$. then $a=frac{x^2}{by^2}$, and $sqrt frac{a}{b} = sqrt{frac{x^2}{b^2y^2}}=frac{x}{by}=frac{n}{m}$.