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!
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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