MHB Solve a Problem & Impress Friends: Proving a/(b+c) for Integers a, b, c

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The discussion centers on proving that if an integer a divides both integers b and c, then a also divides the sum b + c. The proof begins by defining that if a divides b, there exists an integer k1 such that b = ak1, and similarly for c with k2. By combining these, it shows that b + c = a(k1 + k2), indicating that a divides b + c. The intuitive explanation emphasizes that since b and c are multiples of a, their sum must also be a multiple of a. This proof effectively demonstrates the divisibility property, leading to an impressive conclusion.
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I'd love to solve this problem and stun a friend...
Prove if a/b and a/c then a/(b+C)
let a,b,c (element of) (integers.)

I'm sure there's a genius that can figure this out.

Thanks!
 
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I take it that "a/b" means "a divides b". Then, if $a$ divides $b$, then there exists an integer $k_1$ such that $b = ak_1$. Similarly, if $a$ divides $c$, then there exists an integer $k_2$ such that $c = ak_2$. Then $b + c = ak_1 + ak_2 = a(k_1 + k_2) = ak_3$ for $k_3 = k_1 + k_2$, and so $a$ divides $b + c$.

More intuitively, if $a$ divides both $b$ and $c$, that means that $b$ and $c$ are both multiples of $a$ (by definition). So their sum must also be a multiple of $a$, and the result follows.​
 
Thanks for the help!
You should have seen the look on his face! Priceless!
 
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