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$m, n \in \mathbb{Z}$ $m^2 = 2n^2 \implies m = 2k$ for some $k \in \mathbb{Z}$

In other words, the first statement implies $m$ is divisible by 2. Why?

My professor used this without proving it.

My idea is:

$m = 2(\frac{n^2}{m})$, but this is not enlightening, since $\frac{n^2}{m} \in \mathbb{Q}$, and it must be an integer for divisibility.

Another idea is:

Since $\mathbb{Z}$ has closure under multiplication, we can say this implies that $m^2$ is divisible by 2, but again, the same issue arises.

Bill Dubuque
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user129393192
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    Does this answer your question? Can I prove (if $n^2$ is even then $n$ is even) directly?, If prime $p \mid ab$, then $p \mid a$ or $p \mid b\ $. – Robert Lee Oct 25 '23 at 03:57
  • How is that even a remote duplicate? The contents of the questions and the questions themselves are suitably different. @RobertLee and I think this deserves a question and an answer. I do not, nor am I remotely close, to the contents of either of those questions in my course and am asking a simple approach on how to tackle a very specific problem. – user129393192 Oct 25 '23 at 04:23
  • Please peruse the linked dupes for all the common proofs of this FAQ. If you have specific questions about these proofs then ask them in comments on the answers. If you get no suitable reply then post a new question specifying precisely which step of the proof that you don't understand. Please always search for answers before posting questions. See here for how to search. – Bill Dubuque Oct 25 '23 at 04:56
  • I don't think you get the point. I didn't recognize this problem as if $n^2$ is even, prove $n$ is even. That's why the title wasn't what you editted it to be originally. @BillDubuque – user129393192 Oct 25 '23 at 21:58
  • @user129393192 If that was your original question, you may edit the title and replace both “even” instances by “divisible by $2$”. – peterwhy Oct 25 '23 at 22:58

1 Answers1

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Hint: What do you get when you multiply two odd integers together?

X Stanton
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  • An odd integer? – user129393192 Oct 25 '23 at 04:25
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    Yes. So what does that tell you? Can m be odd? – X Stanton Oct 25 '23 at 04:27
  • Oh, $m^2$ is even. But what does that tell us? – user129393192 Oct 25 '23 at 04:29
  • Ok, let us assume from there we say that $m$ must be even, via something I haven't proved. I see. Then we have it's divisible, right? – user129393192 Oct 25 '23 at 04:29
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    Correct. To show that $m$ is even, just ask yourself this. If $m$ is odd, can $m^2=2n^2$ as described? – X Stanton Oct 25 '23 at 04:33
  • If you're implying a proof by contradiction, how would I show that $m^2$ must be odd and therefore we can't have that it equals $2n^2$? – user129393192 Oct 25 '23 at 04:37
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    An odd number times an odd number is odd. So if $m$ is odd, then $m^2$ is odd. But $m^2=2n^2$ meaning $m^2$ is a multiple of $2$. It follows that $m^2$ can't be odd. Getting to the conclusion that $m$ is even should be a cakewalk now. – X Stanton Oct 25 '23 at 04:38
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    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing). – Bill Dubuque Oct 25 '23 at 04:40
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    How is that best for site health @BillDubuque?!?! I literally did not know how to proceed without his help, and it wasn't helpful for me to get this closed down with no answer, since I had a legitimate question. Just because it was an application of a "dupe", doesn't make it a dupe. – user129393192 Oct 25 '23 at 04:43
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    So basically, we know if $m$ is odd then $m = 2k + 1$, $m^2 = (2k + 1)^2 = 2(2k^2 + 2k) + 1$ so it is odd. But $m^2 = 2n^2$, so by definition it must be even. Contradiction. @XStanton – user129393192 Oct 25 '23 at 04:46
  • $m^2=2n^2$, so $m^2$ is even. You should be able to fill out the remaining part. – X Stanton Oct 25 '23 at 04:48
  • I think I got it in the comment @XStanton I originally had just mistyped $m$ for $m^2$ in the last "even" statement. – user129393192 Oct 25 '23 at 04:49