If $n^2$ is odd, is $n$ odd?

Question

I don't really know how to solve the question: if $n^2$ is odd, is $n$ odd? I think it might be odd, because every equation I've done such as $5^2$, $3^2$, etc. End up as odd numbers, is the way I go about answering this question wrong?

Assume $n$ is not odd, then $n$ is even. Then $n^2$ would be even. Contradiction the original statement that $n^2$ is odd. Also, please use latex front. — Frank, Dec 18 '23 at 00:20
@Frank That's not really contradiction, that's contraposition with a contradiction shoe-horned in at the end. — Arthur, Dec 18 '23 at 00:37
Duplicate of Prove that if n is even then $n^2$ is even and if n is odd then $n^2$ is odd? — Martin Brandenburg, Dec 18 '23 at 00:46
@Frank This is an answer to the question, not just a comment. Comments should only be used to clarify, not answer the question. See How do comments work for more information. — Martin Brandenburg, Dec 18 '23 at 00:48
@Blue This is an answer to the question, not just a comment. Comments should only be used to clarify, not answer the question. See How do comments work for more information. — Martin Brandenburg, Dec 18 '23 at 00:48

score 1 · Answer 1 · answered Dec 18 '23 at 00:39

Yes, the way you go about answering the question is wrong for two reasons.

First reason is that there are infinitely many odd numbers, and checking them one by one you'll never be able to finish your proof.

Second reason is that you are checking wrong thing, you are picking an odd number $n = 3,5$ or some other and then checking if $n^2 = 3^2,5^2$ is odd, but you are asking: if $n^2$ is odd, is $n$ odd, the other direction.

You have to ask yourself this: can you find even $n$ such that $n^2$ is odd? The answer is no, because product of two even numbers is always an even number. Therefore, if $n^2$ is odd, so must be $n$.

Samyog Poudel · Answer 2 · 2023-12-18T01:48:07.280

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Yes. If n^2 is odd then n is odd.

This statement can be proved by proving its contrapositive i.e. p implies q is equivalent to ~q implies ~p.

So we can show if n is even then n^2 is even.

Proof:

If n is even then it can be expressed in the form n=2m for some positive integer m.

Now, n^2 = (2m)^2

     = 4m^2
 =2(2m^2)

 =2k where, k=2m^2 is a positive integer.

Thus n^2 is also even.

So, n is even implies n^2 is even proves that n^2 is odd implies n is odd.

edited Dec 18 '23 at 01:48

answered Dec 18 '23 at 01:46

Samyog Poudel

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Welcome to stackexchange. That said, please do not rush to answer questions where the OP has shown no work of their own. When you do ask or answer, use mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker Dec 18 '23 at 01:53
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^Answers help everyone reading these questions, not just the OP. Any correct answer should be submitted. – John Dec 18 '23 at 01:59

If $n^2$ is odd, is $n$ odd?

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