8

I'd like to prove that if ${m}^{2}$ is a multiple of $3$, then ${m}$ is also a multiple of $3$. Similarly, I'd like to disprove that if ${n}^{2}$ is a multiple of $4$, then ${n}$ is also a multiple of $4$.

Per the comment from @thisismuchhealthier, the context is that I'm studying the proof of the elementary theorem from analysis that there is no rational number whose square is $2$ and the related statements that $\sqrt{3}$ and $\sqrt{6}$ are both irrational, but there is a rational number whose square is $4$.

InsideOut
  • 6,883
Ursus Frost
  • 203
  • 3
    Hint 1: 3 is prime, Hint 2: the second requires only a counterexample – jxnh Jul 20 '14 at 20:42
  • Since 3 is prime, then it seems very obvious to me that $m^2$ is or is not a multiple of 3 just the same as $m$ is or is not. 4, on the other hand, is composite, so that if $n \equiv 2 \mod 4$, then $n^2 \equiv 0 \mod 4$ but that also holds when $n \equiv 0 \mod 4$ to begin with. – Robert Soupe Jul 21 '14 at 03:49

7 Answers7

9

Hint $\ 3\mid (m\!-\!1)m(m\!+\!1)=\color{#c00}{m^3\!-m},\ $ so $\ 3\mid\color{#0a0}{m^3}\,\Rightarrow\, 3\mid \color{#0a0}{m^3}\!-(\color{#c00}{m^3\!-m}) = m$

Bill Dubuque
  • 272,048
  • You keep surprising me with strange solutions. – Bart Michels Jul 20 '14 at 21:07
  • @BillDubuque Where is barto's answer? – hola Jul 20 '14 at 21:14
  • 1
    @BillDubuque Your solution is awesome! How you got this strange idea? – hola Jul 20 '14 at 21:15
  • @pushpen.paul My ghost-answer re-appeared. – Bart Michels Jul 20 '14 at 21:16
  • It's not strange at all. More like a little trick. Nice solution anyway. – Deathkamp Drone Jul 20 '14 at 21:17
  • @barto It is not clear what do you mean by My ghost-answer – hola Jul 20 '14 at 21:17
  • 3
    In fact you make it even "simpler" by assuming $3\nmid m$. Then $3\mid(m-1)(m+1)=m^2-1$ and since $3\mid m^2$, we would have $3\mid m^2 - (m^2-1)=1$, absurd. – Deathkamp Drone Jul 20 '14 at 21:18
  • @DeathkampDrone This is is also very elementary and great! This is the advantage of such sites, even a very basic problem can be seen in variant ways. – hola Jul 20 '14 at 21:19
  • 1
    @DeathkampDrone Right, but my goal was to give an example of a "direct" proof, i.e. not by contradiction. – Bill Dubuque Jul 20 '14 at 21:22
  • I'm not saying my proof is better than yours, in fact I prefer yours since as you said it does not resort to proof by contradiction. I was just showing how one can devise different tricks to solve elementary problems. – Deathkamp Drone Jul 20 '14 at 21:24
  • 1
    @DeathkampDrone Both proofs are trivially equivalent (even an automated theorem prover can rewrite one to the other). Generally it's better to avoid contradiction when it is not needed since doing so is often constructively fruitful (e.g. consider Euclid's proof of infinitely many primes). – Bill Dubuque Jul 20 '14 at 21:28
  • The only reason I posted it is because it involves $m^2$ and not $m^3$, which at least in my opinion makes it nicer given the initial information we had (and kind of justifies the proof by contradiction). Anyway, I hope you don't mind me posting the same proof written in a slightly different manner. – Deathkamp Drone Jul 20 '14 at 21:34
  • 1
    @DeathkampDrone Your comments are quite welcome! The point of my reply was simply to explain why I chose the direct way vs. the more common proof by contradiction. Generally, I like to stimulate reasoning "outside of the box". – Bill Dubuque Jul 20 '14 at 21:42
  • @BillDubuque This answer looks very elegant and I have no doubt that it is correct and finding a direct proof is impressive, unfortunately I am too ignorant to understand it and have thus accepted another answer. Perhaps some day in the very distant future I'll know enough to understand it. – Ursus Frost Jul 21 '14 at 00:54
  • @Ursus Please accept the answer you find most helpful. I can explain the above further if you tell me what is not clear to you. The first divisibility comes from the fact that one of $3$ consecutive integers must be divisible by $3$, hence so is their product. The final divisibility is true because multiples of $3$ are preserved under differences: $,3a-3b = 3(a-b).\ \ $ – Bill Dubuque Jul 21 '14 at 01:01
  • 1
    @BillDubuque As a demonstration of my ignorance, I was unaware that the product of 3 consecutive integers is divisible by 3. However, I just read a couple of inductive proofs and it makes sense, so thanks for enlightening me. As I read your answer, I think I see that it proves that if ${m}$ is divisible by 3, then ${m}^{3}$ is also divisible by 3. However, I guess it still escapes me how it directly proves that this also means that ${m}^{2}$ is divisible by 3 which I assume is the implication. I don't doubt the validity as I've tested it on several multiples of 3, just don't understand it. – Ursus Frost Jul 21 '14 at 01:51
  • 2
    @Ursus $ $ Great! $ $ Suppose that $,3\mid m^2.,$ Then $,3\mid m^3 = m\cdot m^2.,$ By my proof, $,3\mid m^3\Rightarrow,3\mid m,,$ completing the proof. In your prior comment you seem to have read the implication in the wrong direction. Please feel welcome to ask further questions if need be. – Bill Dubuque Jul 21 '14 at 02:05
  • @BillDubuque Thanks! You are both correct that I was reading the implication the wrong way. It now makes sense. – Ursus Frost Jul 21 '14 at 02:48
7

If $m = 3k+1$, then $$ m^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1 $$ and if $m = 3k+2$, then $$ m^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k+1) + 1 $$ Thus, if $m^2$ is a multiple of $3$, then $m$ is a also a multiple of $3$.

(because if $m$ is not a multiple of $3$, then neither is $m^2$)

Community
  • 1
Mathsource
  • 5,393
2

First part:

Consider three cases where $m$ is of the form $3k, 3k+1$ and $3k+2$ respectively. Squaring, you'll get only the first one (of the form $9k^2\equiv3k^2$) is the multiple of $3$ (the other two are $9k^2+6k+1=3(3k^2+2k)+1\equiv3k^2+1$ and $9k^2+12k+4=3(3k^2+4k)+1\equiv3k^2+4 \equiv3k^2+1$). Now convince yourself that for $m^2$ to be a multiple of $3$, $m$ must be of the form $3k$, i.e., $m$ is a multiple of $3$.

I put $\equiv$ signs as when dividing by $3$ these expressions are equivalent.

Other (elegant) way to state this problem is: Since $3$ is a prime and $3 | m^2=m\times m$, hence $3$ must be a factor at least one the factors of $m^2$; i.e. $3 | m$.

Second part:

As the wise people say: One counter-example is sufficient to disprove, take any $n$ which even and not a multiple of $4$ (like, $2$, $6$, $10$ etc). Clearly, $2^2=4$ is a multiple of $4$, but $2$ is not.

I have created one follow-up question here.

Community
  • 1
hola
  • 1,339
  • Thanks for the answer. It is helpful. However, I believe if I follow your logic for multiples of 6--which is not a prime number--then I reach the same conclusion. That is if ${m}$ is a multiple of 6, then ${m}^{2}$ is also a multiple of 6. Thus, it seems that the defining characteristic of 3 is not that it is prime but something else. What am I missing? – Ursus Frost Jul 21 '14 at 00:48
  • @UrsusFrost If $m$ is a multiple of $6$, then $m^2$ will be as well. It is true. What makes you think that it is wrong? – hola Jul 21 '14 at 03:49
  • 1
    @UrsusFrost Further, if $m$ is a multiple of $6$, $m^2$ is a multiple of $36$ as well. – hola Jul 21 '14 at 03:50
  • @symplectomorphic I believe that prime, composite is not the correct characterization. It should be perfect square and not perfect square. – hola Jul 21 '14 at 04:06
  • @UrsusFrost You are probably asking the reverse way: if $m^2$ is a multiple of $6$, then will $m$ be as well? Yes, as $6=3\times 2$ is not a perfect square. – hola Jul 21 '14 at 04:10
  • I have changed this characterization to perfect square and not a perfect square . – hola Jul 21 '14 at 04:24
  • 1
    It's not sufficient to just exclude perfect squares, the proper condition is that if $n$ is squarefree (that is, has no repeated prime factors) then $n \mid m \Leftrightarrow n \mid m^2$. – jxnh Jul 21 '14 at 04:49
1

First proof (I asume $m \in N$, is natural):

$m$ can be written as a product of different powers of prim numberts $p_i$ $m = p_1^{n_1}*p_2^{n_2}*...$

So $m^2 = p_1^{2n_1}*p_2^{2n_2}*... = 2^{2*n_1}*3^{2n_2}*...$. Thus of $m^2$ is a multiple of $3$ it must be $n_2 >0$. Therefore $p_1^{n_1}*p_2^{n_2}*...$ is also a multiple of $3$.

Second disproof: By counter example $n^2 = 4$ so it is a mutliple of 4. But $n = \pm \sqrt 4 = \pm 2$ is not a multiple of $4$

user347834
  • 3
Matthias
  • 2,166
1

We prove the contrapositive: If $3\nmid m$ then $3\nmid m^2$.
If $3\nmid m$, then $\gcd(3,m)=1$ because $3$ is prime. By Bézout's theorem, $mx+3y=1$ for integers $x,y$.
Squaring yields $m^2x^2+3(3y^2+2mxy)=1$, that is, $m^2X+3Y=1$ hence $\gcd(m^2,3)=1$.

Note: this solution does not rely on FTA or the uniqueness of quotient and remainder, but does extend to all primes. Correction (Thank you, Bill): The common proof of Bézout's theorem does rely on uniqueness of quotient and remainder.

Bart Michels
  • 26,355
  • 1
    One easily extends the Bezout-based proof to show that prime $,p\mid ab,\Rightarrow,p\mid a,$ or $,p\mid b,,$ and this immediately yields uniqueness of prime factorizations. So the above is essentially using a special case of uniqueness of prime factorizations. Further the proof of Bezout's identity for the gcd does (essentially) depend on the division algorithm (i.e. that $,\Bbb Z,$ is a Euclidean domain). – Bill Dubuque Jul 20 '14 at 21:08
  • Oh yes sure, wasn't watchful when writing this, didn't have the proof in mind. I'll delete it because this makes it have no more value than other answers. – Bart Michels Jul 20 '14 at 21:09
  • No need to delete, as its redundancy has importance still. – jiten Dec 04 '18 at 01:22
0

Hint: To prove that if 3 divides $n^2$ then it divides $n$ you can prove that if 3 doesn't divide $n$, it doesn't divide $n^2$.

hola
  • 1,339
Scientifica
  • 8,781
0

if $n$ is not a mutiple of three (it is $1$ or $2$ $\mod 3$), then $n^2$ is $1^2=1$ or $2^2=1$ $\mod3$ none congruent to three. for the second question, look at $2$ and $2^2=4$

cirpis
  • 1,868