Prove that if $a$ and $b$ are odd, coprime numbers, then $\gcd(2^a +1, 2^b +1) = 3$

Solution 1:

Note that both $2^a+1$ and $2^b+1$ are odd, so any common divisor is odd.

Now, if $d$ divides $2^a+1$ and $2^b+1$, then it also divides $(2^a+1)-(2^b+1) = 2^a-2^b$, hence it divides $2^{\min(a,b)}(2^{|a-b|}- 1)$. But since $d$ is odd, then $d$ divides $2^{|a-b|}-1$. That is, if, say, $a\geq b$, then $$d|2^{a}-1,2^b-1 \Longleftrightarrow d|2^a-1, 2^b-1, 2^{a-b}-1.$$ But if $d$ divides both $2^{a-b}-1$ and $2^b-1$, then it divides $$2^b(2^{a-b}-1) + (2^b-1) = 2^a-1.$$ So: $$d|2^a-1,2^b-1 \Longleftrightarrow d|2^b-1, 2^{a-b}-1.$$ We are assuming $a\geq b$; be careful with that assumption; more generally, we have: $$\text{For }a,b\text{ odd, }\gcd(2^a-1,2^b-1) = \gcd(2^{\min(a,b)}-1, 2^{|a-b|}-1).$$

This looks very much like the gcd identity $$\gcd(x,y) = \gcd(y,x-y),$$ which is very useful. Maybe you can make it work for you?

Solution 2:

A very useful fact: If $a$ and $b$ are relatively prime, there exist integers $x$ and $y$ such that $ax+by=1$.

We can arrange for $x$ to be $\ge 0$ and $y$ to be $\le 0$. So there are non-negative integers $u$ and $v$ such that $au=bv+1$. Thus $$2^{au}=2^{bv+1}=2\cdot 2^{bv}. \qquad\text{(Equation 1)}$$

Let $m$ be any common divisor of $2^a+1$ and $2^b+1$. Then $2^a \equiv -1\pmod{m}$ and $2^b\equiv -1 \pmod m$. It follows that $$2^{au} =(2^a)^u \equiv (-1)^u \pmod{m} \qquad\text{and}\qquad 2^{bv} =(2^b)^v \equiv (-1)^v \pmod{m}.$$ From Equation $1$, we conclude that $$(-1)^u \equiv 2\cdot (-1)^v\pmod{m}.$$ Since $a$ and $b$ are odd, $u$ and $v$ have opposite parity. Thus $-1\equiv 2\pmod {m}$, that is, $3\equiv 0 \pmod m$, and therefore $m$ divides $3$.

If $c$ is odd, then $2^c\equiv -1\pmod{3}$, so $3$ divides $2^c+1$. Thus $3$ divides our gcd. But our gcd divides $3$, so it is $3$.