No irreducible element in a UFD can be a square. 
We wish to prove that no irreducible element in a Unique Factorization Domain $D$ can be written as a square of some element in its field of fraction $F$. 

Take $x$ in $D$, and we go for a contradiction. Write $x=\left(\frac{a}{b}\right)^{2}$. My goal is, of course, to contradict the irreducibility of $x$. But, I'm having some trouble.
 A: You can assume $a$ and $b$ coprime. We get $xb^2 = a^2$. Now a prime factor of $a$ must divide $x$ 2 times, thus $x$ is not irriducible. Contradiction.
A: As hinted, the proof is virtually the same as proving the square root of 2 is irrational. We can assume by dividing out a power of $x$ that at least one of $a,b$ is not divisible by $x$. Then since
$$b^2x=a^2$$
and we are in a UFD, $x\mid a$ (because $x$ is prime). Write $a=cx$ where $c$ is in our base ring. Then
$$b^2x=c^2x^2$$
so
$$b^2=c^2x$$
and hence $x$ divides $b$, a contradiction.
A: Hint $\ $ In $\,a^2 = x b^2$ the prime $x$ occurs to even power on LHS but odd power on RHS, contradiction. Note the the proof uses the existence and uniqueness of prime factorizations.
Remark $\ $ More generally the classical proof of the Rational Root Test easily generalizes to UFDs. In particular it implies that if $\,z^2 - d\, $ has a rational root (i.e. in the quotient field of $D)$ then that root is integral, i.e. in $D,\,$ just like the classical case $D = \Bbb Z$
