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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 $a$.2011-11-13
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    Assuming only integers allowed how about $4^3=8^2$...2011-11-13
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    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.2011-11-13
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    Try: $n$ even $\Rightarrow$ $m$ even $\Rightarrow$ $m^3$ divisible by 8 $\Rightarrow$ $n^2$ divisible by 8 $\ldots$2011-11-13
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    If I take n as 2 though, then isn't this not necessarily true? Or is it due in the part that I am constrained by $m^3=n^2$ that I cannot use this as a counterexample?2011-11-13
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    A cube us equal to a square precisely when it's a sixth power. For example $2^6=64$, so $64=8^2 =4^3$.2011-11-13
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    @Joel: yes, of course, my apologies.2011-11-14

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