Proving GCD's are equal

Question

Let the non-zero elements $a,b \in \mathbb{Z}$. Show that for elements $k,l \in \mathbb{Z}$ where $b = la+k$, $gcd(a,b) = gcd(a,k)$.

What's a good way to start thinking about this proof?

Answer 1

Note the sixth listed property of the Properties section in the "Greatest common divisor" Wikipedia article gives

If $m$ is any integer, then $\gcd(a + m \cdot b, b) = \gcd(a, b)$.

Using this you have, by switching the $a$ and $b$ above and using $m = -l$ as being any integer, that

$$\gcd(a,b) = \gcd(a, b - la) = \gcd(a, k) \tag{1}\label{eq1A}$$

Answer 2

Hint: Use https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity

So we have $Aa+Bb=g$, now sub $b=la+k$ and rearrange to $A'a+Bk=g$.