Anonymous
Anonymous asked in Science & MathematicsMathematics · 3 weeks ago

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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  • mizoo
    Lv 7
    3 weeks ago
    Favourite 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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