0

Here was my take on the proof. We already know that since $(a,b)=1$, there exist integers $x,y$ such that $ax+by=1$. Let $d=(a+b,a-b)$. Then $d|(a+b)$ and $d|(a-b)$. In particular, there exist integers $k_1, k_2$ such that $$a+b=dk_1$$ $$a-b=dk_2$$ Thus depending whether we add or subtract the two equation we get both $$2a=d(k_1+k_2)$$ $$2b=d(k_1-k_2)$$ Thus, $d|2a$ and $d|2b$. But then also $$d|(2ax+2by)\Rightarrow d|2(ax+by)$$ Thus, $d|2$ and thus $d=1$ or $2$.

I don't know why, but I feel like there is a problem with my argument. Perhaps something I haven't said...I feel like it's almost there, but missing some sort of justification, but it's late and I can't figure it out....

Lalaloopsy
  • 1,973

1 Answers1

2

No. There is no problem. But you can "encapsulate" your proof using rules like $$a|b,a|c\Rightarrow a|(b,c),a|mb+nc$$ So by $$d|a+b,d|a-b$$ you directly get $$d|2a, d|2b$$ and then $$d|(2a,2b)=2(a,b)=2$$

Examples for both cases exist: $$(5,3)=1, (5-3, 5+3)=(2,8)=2$$ $$(4,3)=1, (4-3, 4+3)=(1,7)=1$$

Colliot
  • 954
  • 5
  • 16