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

Martin Sleziak
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Jacob
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3 Answers3

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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 $a=b=1$....

N. S.
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  • See my answer for a simple proof that works much more generally. – Bill Dubuque Oct 07 '12 at 07:34
  • @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 07 '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 07 '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. – GeoffDS Oct 12 '12 at 20:12
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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 $\rm\:b/a = x = c/d\in\Bbb Z.$

Bill Dubuque
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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 07 '12 at 09:49
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    @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 07 '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 07 '12 at 12:01
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    @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 07 '12 at 12:11
  • @Garry Myerson The link you gave says so! – PAD Oct 07 '12 at 12:47
  • In any case, the rational root is specific here. It is $\frac{b}{a}$. Therefore the denominator which is $a$ should divide the leading term which is $1$. – PAD Oct 07 '12 at 12:53
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    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 07 '12 at 13:00
  • @Pantelis My proof is indeed one of the standard ways to prove that well-known result about square-roots. Generally, integrally-closed (monic Rational Root Test) is weaker than unique factorization, i.e. UFD $\Rightarrow$ integrally-closed, but the converse fails; e.g. the above proof works in all rings of integers of quadratic number fields $\rm:\Bbb Q(\sqrt{d}),:$ which includes infinitely many non-UFDs. See here for further discussion. – Bill Dubuque Oct 07 '12 at 15:21
  • @Pantelis, I didn't give any link --- what are you talking about? Anyway, the rational root theorem says the rational can be written in such a way that the denominator divides the leading coefficient; it doesn't say that's the only way to write the rational. Would you deny that $42/6$ is a root of the polynomial $x-7$? – Gerry Myerson Oct 08 '12 at 06:34
  • @GerryMyerson I am sorry! I confused you with Bill Dubuque! In his solution he has a link to a wikipedia article on the Rational root theorem. The proof in Wikipedia uses the fact that the rational is reduced. I do not object to $\frac{42}{6}$ being a root of $x-7$ but can it be a root of $x^2-6$? In your example $a=6$ and $b=42$. – PAD Oct 08 '12 at 10:42
  • No, but it can be (and is) a root of $x^2-49$. But I'm not sure that I see what we are discussing. – Gerry Myerson Oct 08 '12 at 11:29
  • @Pantelis 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 $\rm:b/a = x = c/d\in\Bbb Z.$ – Bill Dubuque Oct 12 '12 at 19:56
  • QBill Dubuque o.k. Now I am convinced! – PAD Oct 15 '12 at 20:43
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Any number in the form of $n^6$ can be expressed in the desired form so there will be infinite solutions for $a$ and $b$.

Norbert
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p.g.rajeswara rao
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