Powers of odd numbers

Algebra Level pending

For five integers a , b , c , d , e a, b, c, d, e we know that the sums a + b + c + d + e a + b + c + d + e and a 2 + b 2 + c 2 + d 2 + e 2 a^2 +b^2 + c^2 + d^2 + e^2 are divisible by an odd number n n . Then a 5 + b 5 + c 5 + d 5 + e 5 5 a b c d e a^5 + b^5 + c^5 + d^5 + e^5 - 5abcde is divisible by:

n 2 n^2 n 5 n^5 n n n 4 n^4 n 3 n^3

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1 solution

Priyanshu Mishra
Jun 23, 2019

Consider the polynomial P ( t ) = t 5 + q t 4 + r t 3 + s t 2 + u t + v P(t) = t^5 + qt^4 + rt^3 + st^2 + ut + v with roots a , b , c , d , e a, b, c, d, e . The condition from the statement implies that q q is divisible by n n . Moreover, since

a b = 1 2 ( a ) 2 1 2 ( a 2 ) , \large\ \displaystyle \sum {ab} = \frac {1}{2} (\sum {a})^2 - \frac {1}{2}(\sum {a^2}),

it follows that r r is also divisible by n n . Adding the equalities P ( a ) = 0 , P ( b ) = 0 , P ( c ) = 0 , P ( d ) = 0 , P ( e ) = 0 , P(a) = 0, P(b) = 0, P(c) = 0, P(d) = 0, P(e) = 0, we deduce that

a 5 + b 5 + c 5 + d 5 + e 5 + s ( a 2 + b 2 + c 2 + d 2 + e 2 ) + u ( a + b + c + d + e ) + 5 v \large\ a^5 + b^5 + c^5 + d^5 + e^5 + s(a^2 + b^2 + c^2 + d^2 + e^2) + u(a + b + c + d + e) + 5v

is divisible by n n .

But since v = a b c d e v = -abcde , it follows that a 5 + b 5 + c 5 + d 5 + e 5 5 a b c d e a^5 + b^5 + c^5 + d^5 + e^5 - 5abcde is divisible by n n , and we are done.

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