For five integers we know that the sums and are divisible by an odd number . Then is divisible by:
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Consider the polynomial P ( t ) = t 5 + q t 4 + r t 3 + s t 2 + u t + v with roots a , b , c , d , e . The condition from the statement implies that q is divisible by n . Moreover, since
∑ a b = 2 1 ( ∑ a ) 2 − 2 1 ( ∑ a 2 ) ,
it follows that r is also divisible by n . Adding the equalities 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
is divisible by n .
But since v = − a b c d e , it follows that a 5 + b 5 + c 5 + d 5 + e 5 − 5 a b c d e is divisible by n , and we are done.