If a and b are relatively prime and ab is a square, then a and b are squares. 
If $a$ and $b$ are two relatively prime positive integers such that $ab$ is a square, then $a$ and $b$ are squares.

I need to prove this statement, so I would like someone to critique my proof. Thanks
Since $ab$ is a square, the exponent of every prime in the prime factorization of $ab$ must be even.   Since $a$ and $b$ are coprime, they share no prime factors.   Therefore, the exponent of every prime in the factorization of $a$ (and $b$) are even, which means $a$ and $b$ are squares.

The conclusion you drew from $a$ and $b$  being coprime is not true: 
"Since a and b are coprime, they share no prime factors. Therefore, the exponent of every prime in the factorization of a (and b) are even"
For example $6$ and $5$ are relatively prime, and none of their exponents are even.
Here is a valid proof using some of your reasoning,
Consider the exponents of the primes in the factorization of both $a$ and $b$, since $a$ and $b$ share no common prime factors, their product will not change the value of their original exponents, therefore the only way for all the exponents to be divisible by 2, is if they were originally divisible by 2, which implies they are both squares.
