GCD of exponents

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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



exponentiation greatest-common-divisor




share|cite|improve this question

















  • 1




    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
    2 days ago






  • 2




    What do you mean by "a global formula for $a^n+b^n?$" Why isn't $a^n+b^n$ a "global formula?"
    – saulspatz
    2 days ago










  • 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
    2 days ago










  • "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 math.stackexchange.com/questions/406703/how-to-simplify-an-bn do u know anything for a^n + b^n??
    – Carlson Bimbuh
    2 days ago






  • 1




    Knowing the size of $n$ you're interested in is pretty relevant :) Thanks for editing it into your question :)
    – postmortes
    2 days ago














up vote
0
down vote

favorite
1












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



exponentiation greatest-common-divisor




share|cite|improve this question

















  • 1




    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
    2 days ago






  • 2




    What do you mean by "a global formula for $a^n+b^n?$" Why isn't $a^n+b^n$ a "global formula?"
    – saulspatz
    2 days ago










  • 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
    2 days ago










  • "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 math.stackexchange.com/questions/406703/how-to-simplify-an-bn do u know anything for a^n + b^n??
    – Carlson Bimbuh
    2 days ago






  • 1




    Knowing the size of $n$ you're interested in is pretty relevant :) Thanks for editing it into your question :)
    – postmortes
    2 days ago












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





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



exponentiation greatest-common-divisor




share|cite|improve this 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




exponentiation greatest-common-divisor





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago









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asked 2 days ago









Carlson Bimbuh

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  • 1




    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
    2 days ago






  • 2




    What do you mean by "a global formula for $a^n+b^n?$" Why isn't $a^n+b^n$ a "global formula?"
    – saulspatz
    2 days ago










  • 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
    2 days ago










  • "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 math.stackexchange.com/questions/406703/how-to-simplify-an-bn do u know anything for a^n + b^n??
    – Carlson Bimbuh
    2 days ago






  • 1




    Knowing the size of $n$ you're interested in is pretty relevant :) Thanks for editing it into your question :)
    – postmortes
    2 days ago












  • 1




    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
    2 days ago






  • 2




    What do you mean by "a global formula for $a^n+b^n?$" Why isn't $a^n+b^n$ a "global formula?"
    – saulspatz
    2 days ago










  • 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
    2 days ago










  • "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 math.stackexchange.com/questions/406703/how-to-simplify-an-bn do u know anything for a^n + b^n??
    – Carlson Bimbuh
    2 days ago






  • 1




    Knowing the size of $n$ you're interested in is pretty relevant :) Thanks for editing it into your question :)
    – postmortes
    2 days ago







1




1




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
2 days ago




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
2 days ago




2




2




What do you mean by "a global formula for $a^n+b^n?$" Why isn't $a^n+b^n$ a "global formula?"
– saulspatz
2 days ago




What do you mean by "a global formula for $a^n+b^n?$" Why isn't $a^n+b^n$ a "global formula?"
– saulspatz
2 days ago












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
2 days ago




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
2 days ago












"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 math.stackexchange.com/questions/406703/how-to-simplify-an-bn do u know anything for a^n + b^n??
– Carlson Bimbuh
2 days ago




"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 math.stackexchange.com/questions/406703/how-to-simplify-an-bn do u know anything for a^n + b^n??
– Carlson Bimbuh
2 days ago




1




1




Knowing the size of $n$ you're interested in is pretty relevant :) Thanks for editing it into your question :)
– postmortes
2 days ago




Knowing the size of $n$ you're interested in is pretty relevant :) Thanks for editing it into your question :)
– postmortes
2 days ago















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