If (a,b) = 1 and c|(a+b), show that (a,c) = (b,c) = 1

Question

I am working on this homework problem:

If $\gcd(a, b) = 1$ and $c|(a + b)$, show that $\gcd(a, c) = \gcd(b, c) = 1$.

Hint: Let $d = \gcd(a, c)$ and show that $d|\gcd(a, b)$.

(An Introduction to Number Theory with Cryptography, J. Kraft)

I can solve this without using the hint:

$$(a, b) = 1 => ax + by = 1$$

$$c|(a + b) => ck = a + b => a = ck - b$$

Substituting gives $c(kx) + b(y - x) = 1$

Since both $n = y - x$ and $m = kx$ are arbitrary constants,

$b(n) + c(m) = 1 => (b, c) = 1$

Similarly, $(a, c) = 1$

How would I solve this problem using the hint?

Answer 1

Hint on hint: Let $d=(a,c)$. As $d|c$, we have $d|a+b$ because $c|a+b$. In particular, this means $d|(a+b)-a=b$ so it follows that $d|(a,b)$. With $(a,b)=1$, what can $d$ be?