i suspect that the greatest common divisor is 17

Yes it is and the proof is very ugly, at least mine. The only nice thing about it is actually proving that $17 \nmid 2^{a^2}\cdot 19^{b^2} \cdot 53^{c^2} + 8$. Proof: $2^{a^2}\cdot 19^{b^2} \cdot 53^{c^2} + 8 \equiv 2^{a^2+b^2+c^2}+8 (mod 17)$. We make four cases, by examining $(a^2+b^2+c^2)$ $(mod 4)$ and the one with the actual problem is if $a^2+b^2+c^2=4k+3$ with odd $k$ but that would imply $a^2+b^2+c^2 \equiv 7(mod 8)$ which is absurd. I'm letting someone else type the reasoning behind the fact that only possibility is $17$, since it is gross prime factorization

mine too...

i was finding the orders thsat divides 16 then the primes we get are 3, 5,17 but it is possible to make $(2^{a^2}\cdot 19^{b^2} \cdot 53^{c^2} + 8)$ divisible by 3 or 5 so we got 17 at last, just need to check that the number is never divisible by 17 no need to check the numbers that 16 divides as $(2^{a^2}\cdot 19^{b^2} \cdot 53^{c^2} + 8)$ is divisible by 2

SidVicious wrote: Yes it is and the proof is very ugly, at least mine. The only nice thing about it is actually proving that $17 \nmid 2^{a^2}\cdot 19^{b^2} \cdot 53^{c^2} + 8$. Proof: $2^{a^2}\cdot 19^{b^2} \cdot 53^{c^2} + 8 \equiv 2^{a^2+b^2+c^2}+8 (mod 17)$. We make four cases, by examining $(a^2+b^2+c^2)$ $(mod 4)$ and the one with the actual problem is if $a^2+b^2+c^2=4k+3$ with odd $k$ but that would imply $a^2+b^2+c^2 \equiv 7(mod 8)$ which is absurd. I'm letting someone else type the reasoning behind the fact that only possibility is $17$ i have typed why 17 is the only possibility, 2|16, and 4|16, lead us to the primes 3 and 5, but I have proved that they are not possible, 8|16 too, but no need coz $2^{a^2}\cdot 19^{b^2} \cdot 53^{c^2} + 8$ can be divisible by 2 and there is one way to verify it, is that 2 is a quadratic residue of 17 while 9 is not, so 17 works

Surprisingly it suffices to compute $\gcd\left((2+8)^{16}-1,(1+8)^{16}-1\right)=17$ Basically, factor with difference if squares, and then carefully reduce the primes possible by cross-checking the factorization. Save the big numbers for the end and use Euclid's Algorithm to finish.

true but in contest, I think the useful skills to use is that order divides group size, or group size divides the power, as even if you panick, this is a conventional method that you will never get wrong, for a prime p the group size is p-1