Prove that if $3|mn$, then $3|m$ or $3|n$

Question

I am trying to prove this for integers $m$ and $n$.

I tried to reach prove that $3|m$ by assuming that 3 does not divide $n$, but this is such a basic assumption of mine already that it is hard for me to prove. Could someone help me? Perhaps give me some hints to get me going?

EDIT: Note that this question was an exercise that came before the paragraph on prime factorization (and primes in general).

See also If $p$ is a prime and $p \mid ab$, then $p \mid a$ or $p \mid b$. — Jonas Meyer, Apr 09 '15 at 19:30
This is not a duplicate, because it is only asking about the case $p=3$, without background on general theory of prime numbers available. (Incidentally, here's a third question related to the generalization: http://math.stackexchange.com/questions/847572/proof-that-a-prime-cant-divide-a-multiplication-of-two-reminders-of-it?lq=1) — Jonas Meyer, Apr 09 '15 at 19:39
@JonasMeyer We should probably choose one and mention it in the common questions list. I have found a few others and it seems the one in my first comment is best received (7 answers, all well-received). The first one you mentioned is actually a proof verification so I would not choose that one. It only has 2 answers (Bill's and Bernard's, and they answered the other one as well). — Bart Michels, Apr 10 '15 at 13:59

score 4 · Answer 1 · answered Apr 09 '15 at 19:34

4

Here is a sketch of a contraposition argument. Suppose that $3$ is not a factor of $m$ or of $n$. Then each is either one greater or one less than a multiple of $3$, i.e., there exist integers $j$ and $k$ such that $m=3j\pm1$ and $n=3k\pm1$ (with independent sign choices). Multiply these together and you will have $3(\text{stuff})\pm 1$, so $mn$ is not a multiple of $3$.

answered Apr 09 '15 at 19:34

Jonas Meyer

53,602

Thank you! This works. I hadn't thought of the fact that there are only two possibilities each for m and n (3k+1 and 3k-1). Is there also a way to quickly show that the situations (m=3k+1 and n=3j-1) and (m=3k-1 and n=3j+1) are the same and that you therefore only need to work out one of the two situations? It doesn't save a lot of time, but I simply wonder whether this can be done in a fast way in a proof. – Dasherman Apr 09 '15 at 19:47
@Dasherman: Rather than consider separate cases, you can just leave in the $\pm$ for both throughout, remembering that the sign choices are independent, and find that it works in general. E.g., it can be accepted that $(\pm 1)\cdot(\pm 1) =\pm 1$ without specifying which signs apply. Likewise the "stuff" will have a couple of $\pm$s in it, which don't hurt anything; "stuff" will be an integer regardless of sign choices. – Jonas Meyer Apr 09 '15 at 19:49
This is quite interesting. I have never done arithmetic with $\pm$. Do you have the rules that apply to $\pm$ for some basic opertions (addition, subtraction, multiplication, division, etc)? – Dasherman Apr 09 '15 at 19:54
@Dasherman: Here all we need to know is that $(\pm a)(\pm b)=\pm ab$, which means in this context that if $a$ or $-a$ is multiplied with $b$ or $-b$, the result is $ab$ or $-ab$. You can probably take that for granted, although you can prove the rules about how signs work in multiplication from the basic properties of addition and multiplication of integers. I would only suggest that you make sure the context is clear, which is why I mentioned that the sign choices for $m$ and $n$ are independent; sometimes when two $\pm$s appear in the same expression, it means both $+$ or both $-$. – Jonas Meyer Apr 09 '15 at 20:08

score 3 · Answer 2 · answered Apr 09 '15 at 19:27

3

Try to decompose m and n into their prime factorization.

answered Apr 09 '15 at 19:27

Eric Stchiike

135

Do you know a different way? This exercise comes from before the paragraph about primes (and prime factorization) – Dasherman Apr 09 '15 at 19:35

score 1 · Answer 3 · answered Apr 09 '15 at 19:28

1

Hint: Use the (uniqueness of the) prime factorization of $n$, $m$ and $nm$.

answered Apr 09 '15 at 19:28

NiU

1,163

score 0 · Answer 4 · edited Apr 13 '17 at 12:21

Since $\,3\mid k\iff 3\mid -k,\ $ wlog we may assume $\,n\ge 0.\,$ We use (strong) induction on $\,n.\,$

Our bases cases $\,n = 0,1,2\,$ are true: $\ 3\mid 0\cdot m\,\Rightarrow\, 3\mid 0,\ $ and $\ 3\mid 1\cdot m\,\Rightarrow\, 3\mid m,\,$ and $\,3\mid\color{#0a0}{2 m}\,\Rightarrow\, 3\mid m = 3m-\color{#0a0}{2m}.\,$
Else $\ \color{#c00}{n\ge 3}\ $ and $\,3\mid nm\,\Rightarrow\,3\mid \overbrace{(n\!-\!3)}^{\large \color{#c00}{\ge\, 0}}m\,\overset{\rm induct}\Rightarrow\,3\mid n\!-\!3\, (\Rightarrow 3\mid n)\ $ or $\ 3\mid m\ \ \ $ QED

Remark $\ $ The same proof works for any fixed prime $\,p\,$ (with $\,p\,$ base cases to check). But this method does not suffice to give a uniform proof for all primes. That requires a new idea idea, e.g. the Division Algorithm, Euclidean Algorithm, gcds or lcms, or ideals, e.g. see this proof.

Prove that if $3|mn$, then $3|m$ or $3|n$

4 Answers4