Prove there no integers $x,y$ such that $x+y=100$ and $\gcd(x,y)=3$

Question

I have not encountered a problem like this before.I was thinking that there are 2 methods to solve this. 1. Assume $x+y=100$ and then prove that $\gcd(x,100-x)\neq 3$. 2. Assume that $\gcd(x,y)=3$, and then prove that $x+y$ can never equal $100$ for any $x,y\in R$. I would appreciate a little guidance on the initial setup. Thank you.

Answer 1

You can use your second strategy. Indeed, if $3 = gcd(x,y)$ then $3$ divides both $x$ and $y$ and hence $3$ divides $x+y$.

Answer 2

$\gcd(x,y) = 3 \implies 3 \mid x$ and $3 \mid y$, i.e., $x=3m$ and $x=3n$, where $m,n \in \mathbb{Z}$. This means we need $3m+3n = 100$, which is not possible.