Background: So, just for fun, I was trying to analyze the types of soultion one may recieve from a quadratic equation, the solutions from $\mathbb{Z},\mathbb{R},\mathbb{C}$ was all rather easy, but when it comes to solutions in $\mathbb{Q}$, I tried applying the rational root theorem but that one has a few criteria that wasn't especially well suited for my needs (or rather, I didn't find a way to apply it). So, here's my question:

Given a rational number $a/b$, when is its square root rational?

Please note that I'm not just asking about square roots of integers, but actually any rational number, such as $1/2$.

My attempt We have $Q = a/b : a,b \in \mathbb{Z}$ and thus: $$\sqrt{Q}=\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{ab}}{b}$$ hence we can deduct that if $a=b^{2k+1} \neq 0 : k \in \mathbb{Z}\rightarrow \sqrt{ab} \in \mathbb{Q}$, then the expression as a whole is rational. That is, since $b^{2k+1}\times b = b^{2m} : m,k \in \mathbb{Z}$ then $\sqrt{b^{2m}}=(b^{2m})^{\frac{1}{2}}=b^{m}$ which is clearly rational.

It seems reasonable that this would be a equivalent relation, but I can't figure out how to prove it.

Thanks in advance.

  • 2
    $\begingroup$ It;s clearer, I think, if you assume that $\gcd (a,b)=1$ from the start. then $\frac ab$ is a square if and only if $a,b$ are both squares. $\endgroup$
    – lulu
    Nov 12, 2017 at 18:15

1 Answer 1


If $q$ has a rational square root, we can write is as $\sqrt{q}=\frac{a}{b}$, so $q=\frac{a^2}{b^2}$. In other words, $q$ must have the property that it is the ratio of two square integers. This can be ascertained by looking at $q$ in simplest form, and checking if the numerator and denominator are both perfect squares. Thus, $\frac94$ has a rational square root, but $\frac95$ does not.

I'm not sure this can be expressed neatly by talking about $a$ as a power of $b$. In the case of $\frac94$, we have $9=4^{\log_49}=4^{\log_23}$, but I don't see that leading anywhere very nice.


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