Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was wondering if there were any two integers $a$ and $b$ where $a^3=b^2$.

share|improve this question
Did you try to find any? Say, starting with $a=1$? –  Gerry Myerson Oct 7 '12 at 4:24
Don't forget a = b = 0 –  Hawk Oct 7 '12 at 4:46

3 Answers 3

Hint $\rm\ a=0\!\iff\! b=0.\:$ Else $\rm\:(b/a)^2 = a\in\Bbb Z,\:$ so $\rm\:b/a = n\in\Bbb Z$ via Rational Root Test (RRT). Therefore $\rm\ a = (b/a)^2 = n^2,\:$ so $\rm\:b = an = n^3,\:$ and, indeed, $\rm\:a^3 = (n^2)^3 = (n^3)^2 = b^2\ \ $ QED

Remark $\ $ Note that the proof did not require unique factorization but only the much weaker Rational Root Test, monic-case. Thus the proof generalizes to any integrally-closed domain.

Update (to answer questions in comments) Suppose that $\rm\:x^2 - a\:$ has a rational root $\rm\:x = b/a.\:$ Cancelling $\rm\:gcd(a,b)\:$ we can write $\rm\:x = c/d\:$ in lowest terms. Then RRT implies that the denominator divides the lead coef, i.e. $\rm\:d\:|\:1,\:$ so $\rm\:d=\pm1,\:$ so $\rm\: x = c/d = \pm\, c\in \Bbb Z,\:$ hence $\rm\:b/a = x = c/d\in\Bbb Z.$

share|improve this answer
But the Rational Root Test says that $b|-a$ and $a|1$. So this gives only the solution $a=1,b=1$. The root test assumes that $(a,b)=1$, so we loose many solutions. Am I right? –  PAD Oct 7 '12 at 9:49
@PantelisDamianou The rational root test is being applied to the roots of $x^2-a=0$, with (for each solution) $x=\frac b a$. We get that $x \in \mathbb Z$ ... –  Mark Bennet Oct 7 '12 at 11:23
Exactly. So, $a=1$. But the numerator, which is $b$ should divide the constant term which is $-1$. So, $b=1$. –  PAD Oct 7 '12 at 12:01
@Pantelis, how do you get $a=1$ from knowing $b/a$ is an integer? Isn't $42/6$ an integer? No one said $b/a$ was in lowest terms. –  Gerry Myerson Oct 7 '12 at 12:11
But once you reach $\sqrt{a}=\frac{b}{a}$ you can say: The square root of an integer is rational iff $a$ is a perfect square. Therefore $a=n^2$. Of course, this uses the Fundamental Theorem of Arithmetic. –  PAD Oct 7 '12 at 13:00

Any number in the form of $n^6$ can be expressed in the desired form so there will be infinite solutions for $a$ and $b$.

share|improve this answer

Yes. $a=n^2$ and $b=n^3$, where $n$ is any integer. (for example, $n=2$ yields, $a=4$ and $b=8$).

It is actually easy to prove using the Fundamental Theorem of Arithmetic that these are all solutions.

P.S. I am really surprised that you missed the obvious solutions: $a=b=0$ and $a=b=1$....

share|improve this answer
See my answer for a simple proof that works much more generally. –  Bill Dubuque Oct 7 '12 at 7:34
@BillDubuque Isn't the proof of the RRT based on the fact that $Z$ is an UFD? Otherwise, how can you speak of reduced fractions and gcd? ;) Nevertheless, nice proof. –  N. S. Oct 7 '12 at 15:08
Euclidean $\Rightarrow$ PID $\Rightarrow$ UFD $\Rightarrow$ GCD domain $\Rightarrow$ RRT, integrally-closed, but none of those implications reverse for general integral domains. Hence, for example, the proof I gave works in all quadratic rings of integers, but a proof using unique factorization may not, since such rings generally are not UFDs. See here for further discussion. –  Bill Dubuque Oct 7 '12 at 15:51
@BillDubuque Weird that you would have a solution that works more generally. I'm just kidding and my joke is a compliment. –  Graphth Oct 12 '12 at 20:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.