1. ## Divisibility (gcd) 6

Show that: $\displaystyle ab\neq 0$ and $\displaystyle (a,b)=1 \Rightarrow (a^{2}+b^{2} , a+b)=?$

I don't know..

3. I believe: $\displaystyle d = (a^2 + b^2, a + b) = 1 \text{ or } 2$

Don't have a proof of it at the moment. But here's a start:

Note that: $\displaystyle a^2 + b^2 = (a+b)^2 - 2ab$

Let $\displaystyle d = ((a+b)^2 - 2ab, \ a + b)$.

Since $\displaystyle d\mid (a+b)$ and $\displaystyle d \mid (a+b)^2 - 2ab$, then $\displaystyle d \mid 2ab$

4. Originally Posted by Sea
Show that: $\displaystyle ab\neq 0$ and $\displaystyle (a,b)=1 \Rightarrow (a^{2}+b^{2} , a+b)=?$
Let $\displaystyle (a^{2}+b^{2} , a+b) = d$

Case 1: d|a or d|b

If d|a then d|b (Why?)

Forcing $\displaystyle d = 1$(Why?)

Case 2: d neither divides a nor b

Observe that $\displaystyle (a,b) = 1 = (a,a-b) = (b,b-a)$

$\displaystyle d| (a^2 + b^2) - a(a+b) = b(b-a)$

By case hypothesis, d|b-a which means d|2b. And similarly d|2a.

Now only the following conditions can be true(since (a,b) = 1)

$\displaystyle (2,b) = 1 \Rightarrow d|2 \Rightarrow d = 1,2$
$\displaystyle (2,a) = 1 \Rightarrow d|2 \Rightarrow d = 1,2$

Thus we have showed that d can only be 1 or 2.

Now to show that there are numbers that achieve both gcd:

Choose $\displaystyle a=1,b=2, (a,b)=1, (a^{2}+b^{2} , a+b) = 1$
and
Choose $\displaystyle a=5,b=3, (a,b)=1, (a^{2}+b^{2} , a+b) = 2$

5. Originally Posted by Isomorphism
Case 1: d|a or d|b

If d|a then d|b (Why?)

Forcing $\displaystyle d = 1$(Why?)

$\displaystyle d|a$ and $\displaystyle d|a+b \Rightarrow d|a$ and $\displaystyle d|b \Rightarrow d|(a,b) \Rightarrow d|1 \Rightarrow d=1$
$\displaystyle d|b$ and $\displaystyle d|a+b \Rightarrow d|b$ and $\displaystyle d|a \Rightarrow d|(b,a) \Rightarrow d|1 \Rightarrow d=1$

I can understand.Thanks a million......