If $p$ divides $a+b$ is it true that $p$ divides $a$ and $b$?

Question

My hunch is that this is true. Consider the prime decomposition of $a$ and $b$, then $p$ cannot divide $a+b$ if it does not appear in the decomposition of $a$ and $b$, so it must divide both numbers. Is my sketch correct?

Answer 1

Your assertion is not correct.

You can say that if $p $ divides $a $ then it must divide $b$, because $p$ divides $a+b $ and $a $ so it divides $a+b-a=b $, in a similar way if $p $ divides $b $ then it must divide $a $.

Your assertion is wrong as you can take $p=2$ and $a$, $b $ any two odd numbers, then $p $ divides $a+b $ but it does not divide $a $ nor $b $.

Answer 2

Taking a simple counter example, let $p =5, a=2,b=3$ then $a+b=5\implies p\mid a+b$

But $5\nmid3$ and $5\nmid2$ so the statement fails.

Answer 3

$2$ divides $1+1$. I think that's the simplest counterexample possible.

Answer 4

It Is Not True Try $a = 11 \quad b = 17 \quad p = 7$. I f $p $ divides $a+b$: Then $p$ divides the sum of the remainders of $a $ and $b$ to their euclidean division to $p$.

Or in an other form: $$a\equiv -b\pmod p$$ Nothing more...