Are 2^((n-2)/2) - 1 and 2^(2n) + 1 Relatively Prime?

Naveenks · Jan 30, 2012

Hi, Can anyone help me how to prove that the two numbers 2^((n-2)/2) -1, 2^(2n) + 1 are relatively prime?
Thanks.

dodo · Jan 30, 2012

Hi, Naveenks,
check the question again; if n is odd, the first number is not an integer.

Besides, when n=18, the first number would be 2^8-1 = 255, and the second 2^36+1, and both are divisible by 17.

Naveenks · Jan 30, 2012

Thanks Dodo...I got it that the two numbers are not relatively prime for all values of n...

Are 2^((n-2)/2) - 1 and 2^(2n) + 1 Relatively Prime?

Discussion

Similar threads

412.00.1.12 are relatively prime for all n.

Graduate What are the 2n-1 relations in the Cabibbo and CKM matrices?

Verifying a proof about primes of the form n^2-2

Sum of a series from n=1 to infinity of n^2/(2+1/n)^n

Proving that ##\lim [\sqrt{4n^2 +n} - 2n] = \frac{1}{4}##

Proving $\dfrac{(n-1)^{2n-2}}{(n-2)^{n-2}}<n^n$ for $n\ge 3$

Prove 43 divides 6^(n+2)+7^(2n+1) for all n≥1

Graduate Proving n! + 1 and (n+1)! + 1 are relatively prime

Graduate Is H a Subgroup of N if |H| and |G:N| are Relatively Prime?

Undergrad Proof of Common Divisor of Relatively Prime Integers