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Can someone help me here? I'm completed stuck in this simple problem.

Let a, b, c $\in \mathbb{N} $, check if the statement below is true or false:
if   a|b then (a+b) | (b+c)

Any tips?
What i've done so far:

If a|b, then b=a.c
To prove that (a+b)|(b+c) there must be a j, such as (b+c) = (a+b)*j

  1. (a+b) -> a+a.c

  2. (b+c) -> a.c + c

  3. (a+b)*j =?= (a.c+c)

Oh lord, just tried one case and found out that it is false.
Take a = 3, b = 6, c =2

  1. a|b -> b =a.c -> 6=3*2
  2. (a+b) | (b+c) -> (b+c) = (a+b)*j -> 6+2=(3+6)j -> 8=9j
    (there's no j in N that satifies this equation
Bill Dubuque
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Hesoyam
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    you should tell us which triples a,b,c you have already tested. Welcome to Math Stack. – 311411 Oct 13 '23 at 22:39
  • press "Edit" above; you should include what you've done so far in your main post. – 311411 Oct 13 '23 at 22:50
  • $a\mid b\Rightarrow \color{#c00}a\mid a+b\mid \color{#c00}{b+c}.,$ To get a counterexample it suffices to choose $,b,c,$ with $,\color{#c00}{b+c},$ is coprime to $,\color{#c00}a,$ e.g. $,b+c = 1,$ or $,1+a.,$ Once all is clear please delete the question since we already have too many answers explaining such matters. $\ \ $ – Bill Dubuque Oct 13 '23 at 23:48
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    Same counterexample there works here if we double $,3\mid 6,$ to $,6\mid 12,$ (to force $a\mid b).,$ Counterexamples abound. Please check for such before posting questions. – Bill Dubuque Oct 14 '23 at 01:24

1 Answers1

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I've just tried one case and found out that it is false. Take a = 3, b = 6, c =2

a|b -> b =a.c -> 6=3*2 (a+b) | (b+c) -> (b+c) = (a+b)*j -> 6+2=(3+6)j -> 8=9j (there's no j in N that satifies this equation

Hesoyam
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