$\gcd((p+1)/2,~p-1) = ?$

$p$ is an odd prime

Question

$\gcd((p+1)/2,~p-1) = ?$

$p$ is an odd prime

dan815 · Answer

what is the GCD algorithm

ganeshie8 · Answer

ganeshie8 · Answer

gcd(a, b) = gcd(a-b, b) = gcd(a, b-a)

ganeshie8 · Answer

i have been messing with that 
but no success..

dan815 · Answer

1/2, 2, 1/2, 1

dan815 · Answer

okay
(p+1)/2 = p/2 + 1/2, 
and p-1 = 2*(p/2 - 1/2)

ganeshie8 · Answer

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dan815 · Answer

i think showing this pattern recurring is better

dan815 · Answer

we can show that  all oodds will be 1 and 2 repeating

dan815 · Answer

let p=2n+1

dan815 · Answer

GCD((2n+1)+1)/2,2n-1)

ganeshie8 · Answer

$p$ is given as an odd prime
so..

dan815 · Answer

if we show this happens to all odd numbers then we can atleast say it happens to all odd primes, the patternn of the 1s and 2s might be random

dan815 · Answer

GCD((2n+1)+1)/2,2n-1) = 1 or 2 in fact we can evn do 1 step better by changing n=2k and 2k+1

dan815 · Answer

to show every 2nd add is 2 and every 1st odd is 1

ganeshie8 · Answer

Oh! for odd numbers also it is giving 1 or 2
didn't notice before.. interesting..

dan815 · Answer

right

dan815 · Answer

oops i mean (2n-1)-1

dan815 · Answer

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