Let d = gcd(a, b), prove kd = gcd(ka, kb)

[youvecaughtme]

I feel like I am convoluting this problem. Any help is appreciated! Also, I'm not sure how to rewrite the first part to make it more readable. Any help here is appreciated, too.

Homework Statement

Let a and b be integers and let d = gcd(a, b). If k is a positive integer, prove that
kd = gcd(ka, kb).

HINT: Set d1 = gcd(ka, kb). Argue that kd is a common divisor of ka and kb, so kd
divides d1 . Next, apply Theorem 5 to argue that d1 divides kd.

2. Relevant theorems
Theorem 4: Let a and b be integers and suppose d = gcd(a, b). Then there exist integers
m and n such that d = ma + nb.

Theorem 5: Let a, and b be integers, where a and b are not both zero. Then gcd(a, b)
exists so let d = gcd(a, b). For an integer c there exist integers m and n such that
c = ma + nb if and only if c is a multiple of d.

The Attempt at a Solution

Proof. Let $a, b, d \in \mathbb{Z}$ such that $d=\mathrm{gcd}(a,b)$. Let $k$ be a positive integer and set $d_1=\mathrm{gcd}(ka, kb)=kma+knb \implies d_1=k(ma+nb)$. Because $d|a$ and $d|b$, $d_1=kd(\frac{ma}{d}+\frac{nb}{d})$. Therefore, $kd|d_1$. Using Theorem 5, because $kd$ divides $d_1$, $d_1$ is a multiple of $d$ and therefore $d_1=ra+sb$

From here, I'm not sure where to go. I don't know how to introduce k back into $d_1=ra+bs$

 

[Nino MMP]

and I'm also not sure how to use Theorem 4. Any help is appreciated!Since d = gcd(a, b), there exist integers m and n such that d = ma + nb by Theorem 4. Now consider kd. Since k is a positive integer, kd is also a common divisor of ka and kb. By Theorem 5, we know kd is a multiple of d, so d divides kd. Now consider d1 = gcd(ka, kb). Using Theorem 5 again, since d1 is a common divisor of ka and kb, we know d1 is a multiple of d. We also know that kd divides d1. Therefore, by Theorem 5, d1 must divide kd. Therefore, d1 = kd, and we have proven that kd = gcd(ka, kb).