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Conjecture: If $m^3 = n^2$ and $n$ is even, then n is divisible by $4$.

The proof falls apart from the beginning.

$n$ is even therefore there is a number $k$ such that $n=2k$

$m^3 = n^2$

$m^3 = (2k)^2$

$m^3 = 4k^2$

$4|4k^2$ therefore $4|n^2$

However, I can't think of an example where a cubic is equal to a square. I also ask with hesitation because we have been studying prime numbers and the Euclidean Key theorem as well as other proofs using the Fundamental Theorem of Arithmetic. So, this approach seems out of place for the section of homework that I'm doing.

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Try $m = a^2, n = a^3$ for some $a$. – Qiaochu Yuan Nov 13 '11 at 20:00
Assuming only integers allowed how about $4^3=8^2$... – user12998 Nov 13 '11 at 20:00
Apparently I didn't spend enough time on this one. I'll be stepping back for a bit to think more about it. Thanks for the input. – user17366 Nov 13 '11 at 20:02
Try: $n$ even $\Rightarrow$ $m$ even $\Rightarrow$ $m^3$ divisible by 8 $\Rightarrow$ $n^2$ divisible by 8 $\ldots$ – Jyrki Lahtonen Nov 13 '11 at 20:08
A cube us equal to a square precisely when it's a sixth power. For example $2^6=64$, so $64=8^2 =4^3$. – Michael Hardy Nov 13 '11 at 20:26

Since $n^2 = m^3$ (i.e. $n^2$ is both a perfect square and a perfect cube), the number of factors of 2 in its prime factorization must be a common multiple of 2 and 3.

The smallest such multiple would be zero, but we can discard that possibility on the grounds that $n$ is even.

So, $n^2$ has at least six factors of 2 in its prime factorization. This means that $n$ has at least three factors of 2 in its prime factorization, and so $n$ is divisible by 4 (indeed, it is divisible by 8).

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How should I go about showing "So, $n^2$ has at least six factors of 2 in its prime factorization." Is this drawn from the fact that $n^2=m^3$? – user17366 Nov 13 '11 at 20:15
@arete It's the next smallest number (after 0) that it is a common multiple of 2 and 3. – Austin Mohr Nov 13 '11 at 20:16

There are multiple problems here:

(1) If $n$ is even then $4|n^2$, but not necessarily $n$.

(2) Your "proof" doesn't really use the fact that $m^3 = n^2$.

Also, there are numbers that satisfy $m^3 = n^2$. For instance, $2^6 = 4^3 = 8^2$.

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Well then, that's a start. – user17366 Nov 13 '11 at 19:59
Yes, I suppose I should take the whole "proof" into account. Otherwise, I'm missing important tools. @JyrkiLahtonen's advice was helpful. – user17366 Nov 13 '11 at 20:13

Here is one way to proceed:

Suppose $m^3=n^2$, and $n$ is even. As you say, $n$ must be $2k$ for some integer $k$. Then $m^3=4k^2$, which means in particular that $2\mid m^3$. Now, either by a quick contradiction, or by using Euclid's lemma, it follows that $2\mid m$. (The point here is that $2$ is prime, so if $2\mid m^3=m\cdot m\cdot m$, then it must divide $m$ or $m$ or $m$. The argument by contradiction is that if $m$ is odd, then $m^3$ is also odd by a direct computation, so $2\mathrel{\not|} m^3$.)

If $2\mid m$ then there is an integer (say, $\ell$) such that $m=2\ell$. Then $m^3=8\ell^3$. Then $8\ell^3=4k^2$, so $2\ell^3=k^2$, so $2\mid k$. Then $k=2a$ for some integer $a$, and $n=2k=4a$ is indeed a multiple of $4$.

But it may be worth thinking about this in slightly more general terms. The equation $m^3=n^2$ tells us that the prime factorizations of the two numbers $m^3$ and $n^2$ must coincide. Start with the prime factorization of $m$ and cube it to find the prime factorization of $m^3$. Similarly, start with the prime factorization of $n$ and square it to find the prime factorization of $n^2$. This tells you that there are primes $p_1<p_2<\dots<p_a$ and $q_1<q_2<\dots<q_b$ and exponents $\alpha_1,\dots,\alpha_a$ and $\beta_1,\dots,\beta_b$ such that $$ p_1^{3\alpha_1}\dots p_a^{3\alpha_a}=q_1^{2\beta_1}\dots q_b^{2\beta_b}. $$ (Here, of course, $m=p_1^{\alpha_1}\dots p_a^{\alpha_a}$ and $n=q_1^{\beta_1}\dots q_b^{\beta_b}$.)

The uniqueness of the prime factorization tells us that $a=b$, $p_1=q_1$, $p_2=q_2$, etc, and $3\alpha_1=2\beta_1$, $3\alpha_2=2\beta_2$, etc.

Now: If $3\alpha_i=2\beta_i$, then $2\mid\alpha_i$ (so $m$ is a square) and $3\mid\beta_i$ (so $n$ is a cube), so $n$, being even, must in fact be a multiple of $2^3=8$.

The advantage of this approach, which perhaps looks a bit more complicated than the first one, is that it generalizes to other situations. So one can now quickly conclude by the same argument that if, say, $m^6=n^{17}$, then $n$ is a sixth power (so if $n$ is even, then it is a multiple of $64$, for example). Or, if $m^6=n^{15}$, then $n$ is a square. Etc.

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First try to prove that every number that is both a square and a cube is a sixth power. (It is easier to prove that every sixth power is both a square and a cube.) Then you have $m^3=n^2=\ell^6$. If $4\mid \ell^6$ then $2\mid \ell^3$. Then you can show that $2\mid\ell^3$ only if $2\mid\ell$. That implies that $8\mid\ell^3=n$, so of course $4\mid n$.

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I'm not quite understanding the approach to show that $m^3=n^2=l^6$ . Is there an abstract example that could be shown? This is the part of the proof that I'm not quite understanding. – user17366 Nov 14 '11 at 0:45
Let $m=\ell^2$, $n=\ell^3$. Thus, $m^3=n^2=\ell^6$. – Joel Reyes Noche Nov 14 '11 at 5:22

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