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I am trying to solve this problem, but I am unsure if my proof if sufficient or not. Anyways, here is what I have tried:

So, by using the definition of "divides" I get:

(i): If $a \mid c$, then there exists an integer $k$ such that $ak=c$. Similarly, if $b \mid c$, then there exists an integer $n$ such that $bn=c$.

(ii): Multiplying these two expressions, you get $(ak)(bn)=cc$ or $ab(kn)=c^2$ (which satisfies $(ab)l=c^2$, where $l$ is an integer).

I also have another way of trying to prove it:

(i): By the same definitions as above, you get $a=c/k$ and $b=c/n$.

(ii): By the defintion of divides, you also have $ab(l)=c^2$.

(iii): Inserting $a$ and $b$ into the equation in (ii), you get $(c^2/kn)*l=c^2$ $\rightarrow$ $c^2l/kn=c^2$ $\rightarrow$ $l/kn=1$ $\rightarrow$ $l=kn$.

(iiii): Inserting $l=kn$ into (ii), you get $c^2(kn)/kn=c^2$ $\rightarrow$ $c^2=c^2$, thus $ab \mid c^2$

Any corrections or pointers are appreciated.

Thomas Andrews
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spectr1
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    For your "another way" once you have $a=c/k,\ b=c/n$ you get right away by multiplying that $ab=(c^2)/(kn)$ which shows already that $ab \mid c^2$ – coffeemath Apr 25 '22 at 19:49
  • More generally, if $a\mid c$ and $b\mid d$ then $ab\mid cd.$ – Thomas Andrews Apr 25 '22 at 20:04
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    I don’t understand step $(ii)$ in your second proof. You can’t assume $l$ exists until you’ve proven $ab\mid c^2,$ so what are you trying to say there? – Thomas Andrews Apr 25 '22 at 20:06
  • Essentially in your second proof, you shouldn’t say “you also have…” you should say, “we need …” – Thomas Andrews Apr 25 '22 at 20:07
  • I dislike using fraction notation/division in number theory proofs as a rule. It potentially leaves the domain of integers. It’s not wrong, but I prefer the first proof, even without the sloppy use of language in the second proof. The two proofs are essentially the same, however. – Thomas Andrews Apr 25 '22 at 20:12
  • @ThomasAndrews, Sorry, in step (ii) I am also applying the definition of "divides" on $ab \mid c^2$ – spectr1 Apr 25 '22 at 20:14
  • But you don’t know yet that $ab\mid c^2,$ so you can’t say “we also have…” which means you think you have already proven $ab\mid c^2.$ $(ab)l=c^2$ is what you want, not what you have. @spectr1 – Thomas Andrews Apr 25 '22 at 20:16
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    The second proof also has the disadvantage that it doesn’t work when $a=1, b=2, c=0.$ Then $n=0,k=0,$ so it is not true that $a=c/n,b=c/k,$ since $0/0$ is undefined. Stick with the first proof, or you have to prove the special case $c=0$ separately. – Thomas Andrews Apr 25 '22 at 20:21
  • @ThomasAndrews Ahh I understand, I want to use the basic step to imply the result. So, I should rather use the contructed $a$ and $b$ to imply the result. – spectr1 Apr 25 '22 at 20:21

1 Answers1

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The first proof is fine.

The second proof doesn’t work when $c=0$ and $a,b\neq 0.$ Then you get the problem of $a=0/0.$

You can solve that by doing the case $c=0$ separately, but, as I mentioned in comments, I dislike doing division as a binary operation in beginning number theory problems - it requires potentially exiting the domain of the integers.

Stick with the first proof. The two proofs are basically the same, but the second one requires special cases.

You could also prove it using more elementary lemmas:

For $u,v,w$ integers,

  1. if $u\mid v$ then $uw\mid vw.$
  2. If $u\mid v$ and $v\mid w$ then $u\mid w.$

Then $1.$ with $u=a,v=c,w=b$ we get that $a\mid c$ implies $ab\mid bc$ and $1.$ with $u=b,v=c,w=c$ we get that $b\mid c$ implies $bc\mid c^2.$

Finally $2.$ with $u=ab, v=bc, w=c^2$ shows $ab\mid c^2.$

Thomas Andrews
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