$gcd(2002+2,2002^2+2,\dots,2002^{2002}+2)$

Albert1 · Sep 11, 2016

find: $gcd(2002+2,2002^2+2,2002^3+2,---------------,2002^{2002}+2)$

kaliprasad · Sep 11, 2016

Albert said:

find: $gcd(2002+2,2002^2+2,2002^3+2,---------------,2002^{2002}+2)$

we have $2002^2+2 = (2002^2-4) + 6 = (2002+2)(2002-2)+ 6$
hence $gcd(2002+2, 2002^2+2) = gcd(2002+2, (2002+2)(2002-2) + 6 = gcd(2002+2,6) = gcd(2004,6) = gcd( 6 * 334,6) = 6$
now each of $2002^n+2$ is divisible by 6 because it is even and and $2002$ is $1$ mod 3 and so $2002^n$ is
1 mod 3 and hence $2002^n+2$ is 0 mod 3 so divisible by 6
so GCD of above all that is $gcd(2002+2,2002^2+2,2002^3+2,---------------,2002^{2002}+2)= 6$

$gcd(2002+2,2002^2+2,\dots,2002^{2002}+2)$

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$gcd(2002+2,2002^2+2,\dots,2002^{2002}+2)$

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