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I am trying to prove the statement "Let $n$ be an integer. If $n^2$ is divisible by $3$, then $n$ is divisible by $3$." by using the direct proof technique.

The following is my proof.

Consider the following definition.

Definition 5.1: A nonzero integer $a$ divides an integer $b$ if there is an integer $j$ such that $b = aj.$

Because $3$ divides $n^2$, by Definition 5.1 it follows that

\begin{align} n^2 = 3j \text{ for some $j \in \mathbb{Z}$} \end{align}

By taking the square root on both sides of the above equation, one obtains

\begin{align} n = \pm \sqrt{3j} \text{ for some $j \in \mathbb{Z}$} \end{align}

Since $n$ is an integer, for the $RHS$ of the above equation to be an integer, $j = 3k^2$ for each $k \in \mathbb{Z^*}$.

Hence,

\begin{align} n = \pm \sqrt{9k^2} = \pm 3k \text{ for each $k \in \mathbb{Z^*}$} \end{align}

By Definition 5.1, one can conclude that $3$ divides $n$.

Is the proof correct?

Reference:

Reading, Writing, and Proving: a Closer Look at Mathematics, by Ulrich Daepp and Pamela Gorkin, 2nd ed., Springer, 2011, pp. 52,55.

lone student
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Approxiz
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    I think there's a potential circularity issue: you use the fact that $\sqrt{3}$ is irrational, but the most well-known proof of that depends on this fact. In any case it's a bit of a copout. – Stephen Donovan Apr 18 '21 at 11:25
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    The easiest proof is based on Euklid's lemma for primes $p$ : If $p\mid ab$ , then $p\mid a$ or $p\mid b$. Apply this to the case $p=3$ and $a=b=n$. – Peter Apr 18 '21 at 11:28
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    Another way than the one given by @Peter using heavier artillery is by the prime number decomposition. – nicomezi Apr 18 '21 at 11:29
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    Just a minor typo: "By taking square" should be "By taking square root" – jjagmath Apr 18 '21 at 11:33
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    @Peter In english, the official language of this site, the name is Euclid. – jjagmath Apr 18 '21 at 11:41
  • @nicomezi That's an immediate consequence of Euclid's Lemma, so it is hardly "heavier" artillery. – Bill Dubuque Apr 18 '21 at 18:03
  • So it is not heavier but it is ? @BillDubuque – nicomezi Apr 18 '21 at 18:41
  • Why do you believe "it is"? Uniqueness of prime factorizations follows by a simple straightforward inductive application of Euclid's Lemma. – Bill Dubuque Apr 18 '21 at 18:45
  • Because I believe that using a theorem coming from a lemma that you can use directly and (more ?) conveniently without having to reproof the theorem in some way is using heavier artillery. What is your interpretation on this ? @BillDubuque – nicomezi Apr 18 '21 at 18:49
  • @nicomezi They are equivalent, so either can be used to prove the other. Whether or not one is more convenient will depend on the context. – Bill Dubuque Apr 18 '21 at 18:56
  • Sorry, I do not see the point of such nitpicking, you are now commenting the little thing in the parenthesis to make a point. Have a nice evening, I think we are done. – nicomezi Apr 18 '21 at 18:57

3 Answers3

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Your proof has a problem here

Since $n$ is an integer, for the $RHS$ of the above equation to be an integer, $j = 3k^2$ for each $k \in \mathbb{Z^*}$.

How do you know this?

jjagmath
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Your error was noted in the other answer.

You can apply Proof by Contradiction:

  • If $n=3k-1$, then $(3k-1)^2=9k^2-6k+1.$

This implies,

$$(3k-1)^2 \equiv 1 ~ \text{mod}~3$$

  • If $n=3k-2$, then $(3k-2)^2=9k^2-12k+4.$

This implies,

$$(3k-2)^2 \equiv 1 ~ \text{mod}~3$$

Thus, it must be if and only if $n=3k$.

$$9k^2 \equiv 0~ \text{mod}~3$$

Generalization:

Let, $m$ be a prime number.

If, $n^2 \equiv 0~ \text{mod}~m$ , then if and only if $n=mk, k\in\mathbb Z.$

lone student
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Welcome to MSE! Here's an alternate elementary proof

$$n^2 \equiv 0 \pmod{3} \implies 3 \mid n^2 $$

Thus there's a factor of $3$ in $n^2$. Now it's very trivial that power of any prime in a perfect square is even, and hence power of $3$ in $n^2$ is even which implies that $3 \mid n$ and this concludes the proof. You can even generalize this for any number. Hope this helps.

BooleanCoder
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  • No, it is not "trivial" but depends on a deep result known as the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations) or closely related results. If you don't rigorously justify such claims then there is no way to determine if your proof is correct or not. See here for more on this important matter – Bill Dubuque Apr 18 '21 at 17:58
  • I don't think Fundamental Theorem of Arithmetic is a deep result... You can show it very easly with infinite division $n = a_0b_0 = a_1a_2b_1b_2 = a_1a_2a_3a_4b_1b_1b_3b_4 = ...$ This process must end, therefore you can break $n$ into primes. – 1b3b Apr 19 '21 at 19:06