# Are there any integer solutions to $a^3=b^2$?

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

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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

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 $a=b=1$....

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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

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 $\rm\:b/a = x = c/d\in\Bbb Z.$

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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 $b$.

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