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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 $a=1$?2012-10-07
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    Don't forget a = b = 02012-10-07

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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 $a=b=1$....

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    See my answer for a simple proof that works much more generally.2012-10-07
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    @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.2012-10-07
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    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](http://math.stackexchange.com/a/164589/242) for further discussion.2012-10-07
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    @BillDubuque Weird that you would have a solution that works more generally. I'm just kidding and my joke is a compliment.2012-10-12
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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 $\rm\:b/a = x = c/d\in\Bbb Z.$

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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?2012-10-07
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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 $x \in \mathbb Z$ ...2012-10-07
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    Exactly. So, $a=1$. But the numerator, which is $b$ should divide the constant term which is $-1$. So, $b=1$.2012-10-07
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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.2012-10-07
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    @Garry Myerson The link you gave says so!2012-10-07
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    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 $1$.2012-10-07
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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.2012-10-07
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    @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](http://math.stackexchange.com/a/164589/242) for further discussion.2012-10-07
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    @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 $x-7$?2012-10-08
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    @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 $b=42$.2012-10-08
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    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.2012-10-08
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    @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.$2012-10-12
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    QBill Dubuque o.k. Now I am convinced!2012-10-15
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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 $b$.