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I have to prove that if 3 divides the product of two integers then it divides one of the integers. At this point in my instructors notes, we do not have Euclid's lemma, nor the concept of primes or gcd, nor modular arithmetic. Basically just elementary algebra. One thing we do have is that every integer is of the form $3k$, $3k + 1$, or $3k + 2$. So, I've been doing my proof as such:

  1. If either $a$ or $b$ is of the form $3k$ then we automatically have that 3 divides $a$ or $b$, so we're done in this case, so assume that $a$ and $b$ are not of the form $3k$.

  2. If $a$ and $b$ are both of the form $3k + 1$, then we have that $(3j + 1)(3l + 1) = 3n$, so $9jl + 3j + 3l + 1 = 3n$ and therefore $3l + 1 = 3(k - 3jl - j)$, and therefore 3 divides $3l + 1$.

  3. If $a$ is of the form $3k + 1$ and $b$ is of the form $3k + 2$, then we have that $(3j + 1)(3l + 2) = 4n$, so $9jl + 6j + 3l + 2 = 3n$, and therefore $3l + 2 = 3(k - 3jl - 2j)$, and therefore 3 divides $3l + 2$.

  4. If $a$ and $b$ are both of the form $3k + 2$, then we have that $(3j + 2)(3l + 2) = 3n$, and therefore $9jl + 6j + 6l + 4 = 3k$. Now here is where my issue is located. We can't just split this up nicely like $9jl + 6j + 3l + 2 + 3l + 2 = 3k$, so that $3l + 2 = 3(3jl + 2l + l) + 2$ because then we have this extra 2 here. I've spent an embarrassing amount of time failing to figure out a correct manipulation here to show $3l + 2 = 3m$ for an actual integer $m$. Could I have some help here with how to achieve this, or perhaps a different direction to take? Am I missing something incredibly simple?

Secondary question: I also have to prove this for 5 and 7 instead of 3. This sounds extremely tedious using my current methods. Is there a smarter idea to do this for these cases?

deb
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    $9jl + 6j + 6l + 4 = 3(3jl+2j+2l+1)+1$. – Anne Bauval Mar 17 '24 at 18:32
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    With $5$ instead of $3$ for instance: $5n=(5j+a)(5l+b)=5(jl+al+bj)+ab$ with $-2\le a,b\le2$. If $-2\le ab\le2$ then you are done: $ab=0$ so $a=0$ or $b=0$. If $ab=4$, write $ab=5-1$ hence $5n=5(jl+al+bj+1)-1$. If $ab=-4$, analogous trick. – Anne Bauval Mar 17 '24 at 18:40
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    That's it, thank you very much Anne. – deb Mar 17 '24 at 18:51
  • Yes, the "smarter way" is to use Euclidean descent, i.e. $,p\mid ab\iff p\mid (p\bmod a)b,,$ or $,p\mid ab\iff p\mid \gcd(p,a)b,,$ see the linked dupe. $\ \ $ – Bill Dubuque Mar 17 '24 at 19:26
  • I updated the linked answer in the dupe so it is clearer how it applies here. If anything remains unclear you are welcome to post comments there. – Bill Dubuque Mar 17 '24 at 20:08
  • We can unify your $3$ cases as follows. By division $,a= r+3q,$ for $,r = 0,\pm1$ so

    $$3\mid ab,\Rightarrow, 3\mid(r!+!3q)b,\Rightarrow, 3\mid rb,\Rightarrow, 3\mid b,,\ {\rm by},\ r = \pm 1,,\ {\rm by}\ ,r\neq 0,,\ {\rm by}\ ,3\nmid a\qquad$$

    Here the first reduction step is $,p\mid ab\Rightarrow p\mid (a\bmod p):!b=rb,$ in the linked algorithm. In this case the second reduction step $,p\mid rb\Rightarrow p\mid (p\bmod r)b,$ is not needed since already $|r|=1$ so we infer $,p\mid b.,$

     – Bill Dubuque Mar 17 '24 at 20:52
  • For primes $,p>3,$ we generally need to apply the second step multiple times to reach $,p\mid b,$ (no more than the number of bits in $r$ if we use signed (least magnitude) remainders [as we did above, where we chose remainder $-1$ vs $2$ modulo $3]$ since then each step at least halves the number of bits). In your notation we rewrite $,3k+2 = 3(k!+! 1)-1 = 3j-1,$ to convert a remainder of $,2,$ to $,{-}1,,$ so then we have the $3$ forms $,3k+r,$ for $,r = 0,\pm1,,$ as in the above proof. $\ \ $ – Bill Dubuque Mar 17 '24 at 21:03

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Take, one integer as $3p+2$ & the other as $3q-1$. So,

$(3p+2)(3q-1)=3m$

or, $9pq-3p+6q-2=3m$

or, $9pq+6q-3m=3p+2$

or, $3p+2=3(3pq+2q-m)$

So, $3p+2$ is divisible by $3$.

Rajesh Bhowmick
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Mar 17 '24 at 21:00