Proving properties of Greatest Common Divisors

Question

I have two questions I'm struggling with

1) Suppose that gcd(a, y) = 1 and gcd(b, y) = d. Prove that gcd(a · b, y) = d

I have 1 = ua + vy and d = sb + ty, and I use linear combination to get d*1 = sb*(ua + vy), + ty*1, and after simplifying I get d = (su)ba + (svb)y + ty, I don't know what to do with the ty term

2) Suppose that gcd(b, a) = 1. Prove that gcd(b + a, b − a) ≤ 2

I just don't really know where to start...

Answer 1

First problem: Since $\gcd(a, y) = 1$, there exist integers $u$ and $v$ such that $1 = ua + vy$. Since $\gcd(b, y) = d$, there exist integers $s$ and $t$ such that $d = sb + ty$. If $1 = ua + vy$ and $d = sb + ty$, then

\begin{align*} d & = 1 \cdot d\\ & = (ua + vy)(sb + ty)\\ & = usab + utay + vsby + vty^2\\ & = (us)ab + (uta + vsb + vty)y \end{align*}

What conclusion can you draw?

Second problem: Apply the Euclidean Algorithm to $b + a$ and $b - a$. Observe that

$$b + a - (b - a) = 2a$$

Thus, $\gcd(b + a, b - a) = \gcd(b + a, 2a)$. If $p$ is a prime such that $p|2a$, then $p|2$ or $p|a$. If $p|a$, you can show that $\gcd(a, b) = 1 \Rightarrow p \not\mid b + a$.

Answer 2

For 1, write $b=db'$ and $y=dy'$, with $\gcd(b',y')=1$. Then $ab=dab'$ and $\gcd(ab',y')=1$ because $y'$ does not have prime divisors that divide either $a$ or $b'$. This proves that $\gcd(ab,y)=d$.