# Why is it the case that the common factors of $x$, $y$ are also common factors of $x + y$

Why is it the case that the common factors of $x$, $y$ are also common factors of $x + y$?

For example, $10$, $4$ share factors $2$, $1$. $10 + 4 = 14$ which has factors $2$, $1$

Similarly, $100$, $160$ share factors $1$, $10$, $2$, $4$, and $260$ has factors $1$, $10$, $2$, $4$.

-
if $ds = x$ and $d t = y,$ then $d(s+t) = x+y$ –  Will Jagy Mar 12 '14 at 22:07
$x = k\cdot d,\, y = m\cdot d \Rightarrow x+y = (k+m)\cdot d$ –  Daniel Fischer Mar 12 '14 at 22:07
$$\begin{array}{ccc} d\mid a~\wedge d\mid b & & d\mid(a+b) \\ \Updownarrow & & \Updownarrow \\ a=d\bar{a}~\wedge~b=d\bar{b} & \Rightarrow & a+b=d(\bar{a}+\bar{b}) \end{array}$$ –  blue Mar 12 '14 at 22:19

If $d$ divides both $x$ and $y$, then $d$ divides $x+y$ as well. This is true because if $x = da$ and $y=db$ for integers $a,b$, then $x+y=d(a+b)$.

-