A puzzle that came when I am half awake When I am about to wake up in the morning, a puzzle crept into my mind.It is
when $\sqrt{a}$ and $\sqrt{b}$ are both non-integers where $a,b$ are positive integers is it possible for $\sqrt{ab}$ to be an integer.
My try: Still in half wakeup
If $\sqrt{ab}$ is an integer then $a + b + \sqrt{ab}$ is a positive integer
=> $(\sqrt{a} + \sqrt{b})^2$ is a positive integer 
But I was unable to move forward from here. 
I tried for different combination but didn't found any such pair $a,b$.
Is one exits. How to find it or prove no such pair exist
Edit: I think for every case of $a=b$, this hold good, so I think that cases should be omitted
 A: Just choose $a,b$ so that neither of them are square numbers but $ab$ is a square number. 
Example: $a=3, b=12, \sqrt{ab}=6$
A: The only divisors of $p^2$, where $p$ is prime, are $1, p, p^2$ and thus cannot be a solution.
Thus, choose square of any composite number $q$.
Since a composite number can be written as $q = \prod\limits_{p_i \text{prime}} p_i^{e_i}$ where $e_i \ge 0$ and there is at least one $e_i > 0$.
Suppose, for the sake of simplicity, $q = p_1p_2$. Then, $q^2 = p_1^2 p_2^2$. Now $q^2 = (p_1p_1p_2)(p_2)$ and neither of $p_1p_1p_2$ and $p_2$ are perfect squares.
Now, set $\sqrt{a} = \sqrt{p_1p_1p_2} = \sqrt{\dfrac{q^2}{p_2}}$ and $ \sqrt{b} = \sqrt{p_2} = \sqrt{\dfrac{q^2}{p_1p_1p_2}}$, both of which are not natural numbers, however their product $ = \sqrt{\dfrac{q^2}{p_1^2p_2^2}} = 1$ is a natural number.
This can be generalized in similar way for general $q = \prod\limits_{p_i \text{prime}} p_i^{e_i}$, and thus you have an infinite number of solutions.
A: if you choose a,b such that gcd(a,b) = 1 it never happens..for any other case,given an a you can always find a "b" satisfying the condition.
A: Let $v_p(n)$ be the exponent of the prime $p$ in the factorization of $n$.
Then, $\sqrt{a}$ and $\sqrt{b}$ are both non-integers and $\sqrt{ab}$ is an integer iff $v_p(a)$ and $v_p(b)$ have the same parity for all primes $p$.
In particular, given $a$, you can find $b$ by taking the product of all $p$ such that $v_p(a)$ is odd. You can find infinitely many $b$ by multiplying that product by a square.
