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I came up with the following:

$2|n^2$ implies that $2|n*n$. We proved in class that if $q|b*p$, then $q|b$ or $q|p$. Therefore, if $2|n^2$, then $2|n$ or $2|n$. So, $2|n$ implies $n=2k$, for some $k ∈ Z$. So, $n^2|(2k)^2$, $n^2 = 4k^2$, where $k^2 ∈ Z$.

So, if $2|n^2$, then $4|n^2$ as desired.

What do you all think?

Mettal
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  • Well if $n$ is odd, then $2\nmid n^2$ so you're done with that case. – saulspatz Dec 06 '18 at 04:24
  • Should it be "if $2\mid \color{red}{n}$ then $4\mid n^2$"? – Dave Dec 06 '18 at 04:24
  • See the statement is - if $2|n^2$ then $4|n^2$. You don't need to think about odd $n$. – tarit goswami Dec 06 '18 at 04:31
  • Hi and welcome to MSE. Have you done prime factorisation? Assuming so, what are the properties of the exponents of the prime factors in the every factorisation of a square number? – it's a hire car baby Dec 06 '18 at 18:57
  • Try adding the answer to this to your question: Is this a problem from a text, or from a class? If from a class, which one, and what theorems have they seen recently? If a text, which one? For example, are they in a context of abstract algebra, or number theory, or intro to proofs, or something else? See this discussion: https://chat.stackexchange.com/transcript/message/47930322#47930322 – it's a hire car baby Dec 06 '18 at 19:46
  • @Dave I think not – it's a hire car baby Dec 06 '18 at 20:10

4 Answers4

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If $2|n^2$, $n^2$ is even. Note that $n^2$ is even if and only if $n$ is even (in $\Bbb Z$). So $n=2m $ for some $m \in \Bbb Z$. Hence $n^2=4m^2$.

cqfd
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Suppose not. Then $n^2=2\times (2m+1)$ for some positive integer $m$. But that means, $n^2=4m+2$. Which is a contradiction, as, any perfect square can leave remainder $0,1$ after division by $4$.

Here is a more satisfactory arument. A statement and it's contrapositive statement has same truth value. Here, the contrapositive statement of $2|n^2\implies 4|n^2$ is $4\nmid n^2\implies 2\nmid n$. So, if we can prove the contrapositive statement is true then we are done.

$$4\nmid n^2\implies 2\nmid n\implies 2\nmid n\cdot n $$ So we are sone!

tarit goswami
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If $2|m^2$ then it means its square root $m$ must be even, or else $m$ would be irrational. If it were odd then there would be no $2$ as part of its prime factors, so multiplying it by itself would not create an even number. We can define $n$ such that $2n=m$. Since $m^2=(2n)^2=4n^2$, then we can easily show $4|4n^2$ and therefore $4|m^2$.

Kyan Cheung
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An import ant part of writing any proof is to determine from the outset what you can assume to be true.

If you can assume the fundamental theorem of arithmetic then you have the unique prime factorisation of $n=p_1^{i_2}\cdot p_2^{i_2}\cdot p_3^{i_3}\ldots$

Then the prime factorisation of $n^2=p_1^{2i_2}\cdot p_2^{2i_2}\cdot p_3^{2i_3}\ldots$

Then since every exponent is even, the power of every prime factor of $n^2$ is either $1$ or a power of $p^2$, which generalises the result you need to every prime.

(In $2^3=8$, 2 is the prime, 3 is the exponent and 8 is the power).

it's a hire car baby
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