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# Greatest common divisor?

How to solve:

1. gcd(2, 2^100 - 1) and gcd ((p+q)^2 , p) where p and q prime numbers

### 1 Answer

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- mizooLv 73 weeks agoFavourite answer
1. gcd(2, 2^100 - 1):

2^100 - 1 is an odd number. The GCD of 2 and an odd number is 1.

2. gcd((p+q)^2 , p), where p ≠ q:

for p = 2, p + q is an odd number and the greatest common divisor 2 and an odd number is 1.

for q = 2:

We know gcd(n, n+1), knowing p is prime and with the same logic, gcd(p, p+2) = 1 => gcd((p+q)^2 , p) = 1

for p, q > 2, p + q is an even number and the GCD of an even number and a prime number is 1.

=> GCD of ((p+q)^2 , p) is 1.

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