GCD of exponents

Question

I have been trying to determine a shorter method for calculating $\gcd(a^n + b^n, |a-b|).$ I noticed that, if I were to calculate (using the above formula) say $a = 100$ and $b = 4,$ starting from $1$ and ending at $n$ (a loop), at a certain point, the answer becomes constant. For $a = 100, b = 4, n = 100,$ I create a loop from $1$ to $n,$ and at each point, I apply the formula, the first answer $(n = 1)$ is $8,$ thereafter it is $32$ till when $n$ becomes $100.$ For optimization, I break out of the loop once two equal consecutive numbers are found and the latest number ($32$ here), becomes the answer. Does anyone know a straightforward formula for calculating $\gcd(a^n + b^n, a-b),$ or better still, my primary concern, a global formula for finding $a^n + b^n$

Note: 1<=a,b,n<=10^12

My Python Code for the above explanation follows suite

from functools import reduce
from fractions import gcd

a, b, n = map(int, raw_input().split()); ans = -1

if a == b:
    ans = (a**n) + (b**n)#a-b will give 0 and gcd(x,0) == x

else:
    for j in xrange(n):
        x = gcd((a**n)+(b**n),abs(a-b))

        if x != ans:
            ans = x
        else:
            break
print ans

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Aug 04 '18 at 12:02
What do you mean by "a global formula for $a^n+b^n?$" Why isn't $a^n+b^n$ a "global formula?" — saulspatz, Aug 04 '18 at 12:44
If you are using python, you won't be able to beat the built-in methods of integer arithmetic in source code. Have a look here for starters. — saulspatz, Aug 04 '18 at 12:48
"What do you mean by "a global formula for an+bn?" Why isn't an+bn a "global formula?" Take for example, a^n - b^n can be sipmlified as https://math.stackexchange.com/questions/406703/how-to-simplify-an-bn do u know anything for a^n + b^n?? — Carlson Bimbuh, Aug 04 '18 at 13:17
Since, for large $n$ and $a,b>1$, the smaller term will be $|a-b|$ a faster approach would be to observe that the gcd must be a factor of $|a-b|$. So find those factors, order them in descending order, and see if they evenly divide $a^n+b^n$. The first one that does is the gcd. — postmortes, Aug 04 '18 at 14:02
@postmortes true, but it takes more than ten seconds to compute a^n where a = 1000000 and n = 1000000. Talk less of computing b^n again before summing them up. — Carlson Bimbuh, Aug 04 '18 at 14:33
Knowing the size of $n$ you're interested in is pretty relevant :) Thanks for editing it into your question :) — postmortes, Aug 04 '18 at 14:35

GCD of exponents

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