-2

My professor claims:

If $c\mid(a+b)$ and $c\mid(a-b)$, then $c\mid a$.

How is this correct? I can't prove it, all I achieved is proving that $c\mid(2a)$ which isn't the same.

Angelica
  • 1,963
Dan
  • 27
  • 4
    How is this related to [tag:lagrange-multiplier], to [tag:chinese-remainder-theorem], or to [tag:divisor-sum]? – José Carlos Santos Dec 09 '21 at 16:05
  • @JoséCarlosSantos don't know where these tags came from, sorry :( – Dan Dec 09 '21 at 16:06
  • If $c$ is odd, then $c|2a$ does imply that $c|a$. So, any counterexample must have $c=2d$ (even number) with $d|a$ but $c \not\mid a$. – Geoffrey Trang Dec 09 '21 at 16:07
  • It not correct if $c$ is even and $a$ and $b$ are both odd. (Ex: $c = 2$ and $a=7; b=3$). But if $c$ is odd then it is true by Euclids lemma. And it will always be true that $\frac c{\gcd(c,2)}$ will divide $a$. For example if $14|a+b$ and $14|a-b$ then $14|2a$ and so $7|a$. – fleablood Dec 09 '21 at 16:07
  • 2
    They came from you, of course, since you're the one who typed the question. – José Carlos Santos Dec 09 '21 at 16:09
  • Your professor probably meant to claim (or should have meant to claim) that if $c$ is odd then.,..... $2$ and evenness is a pesky little case-checking possibility that frequently sneaks into these time of factorizing problems. – fleablood Dec 09 '21 at 16:17

2 Answers2

0

It is not true.

Let $c$ be $2$ and $a=b=1$.

Your conclusion that $c|2a$ is correct.

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
0

It's not true as you point out.

But it does mean that $c|2a$ (and $c|2b$). And that means:

  1. If $c$ is odd then $c|a$ and $c|b$.
  2. If $c$ is even then $\frac c2|a$ and $\frac c2|b$ (and maybe $c|a$ [or maybe not]).
fleablood
  • 124,253