Problem: Find $(2^{a}+1,2^{b}+1),$ where $a$ and $b$ are prime numbers greater than $3$ and $(x,y)$ represents the $\gcd(x,y).$
My Attempt: Observe that $(2^a+1,2^b+1)=(2^{a-b}-1,2^b+1)$ if we assume that $a>b.$ Thus it is obvious that $(2^a+1,2^b+1)=2^{(a,b)}\pm1.$ Since $(a,b)=1$, we have $(2^a+1,2^b+1)=1,3.$ I can't figure out the condition under which $(2^a+1,2^b+1)=3.$ Please help.
I will give an alternative proof using digital roots:
A number with digital root $3$ or multiple of $3$ will always be divisible by $3$. Digital roots are those that: $d \equiv n \pmod 9$.
Now powers of $2$ starting from $2^{0}$ to $2^{5}$ follow the sequence: $\{1,2,4,8,7,5\}$. Thus from $2^{6}$ to $2^{11}$ will begin repeating the sequence, and so on.
What do we get from here? Every power of $2$ with an odd number of exponent will follow the sequence $\{2,8,5\}$. Why? Let's explain it:
Remark the sequence beginning from $2^0$ to $2^5 \Rightarrow \{1,2,4,8,7,5\}$ with 0(even) you get 1, with 1(odd) you get 2, with 2(even) you get 4, with 3(odd) you get 8, with 4(even) you get 7 and with 5(odd) you get 5. If you compose the sequence generated by odd exponents then you get $\{2,8,5\}$, as explained above.
Now since you have $2^{a}+1$ with $a$ odd prime, then just sum $1$ to every number of the sequence and u get $\{3,9,6\}$. Since a is odd prime that means that every exponentiation plus 1 will be divisible by 3.
So $2^{p}+1$ with $p$ odd, is always divisible by $3$. That covers prime odd exponents too.