simple question !!!

First , see that  (1 +x + x^2 + x^3 + x^4 )(x-1 ) = x^5-1;

so we can state that ( x-1 ) ( P(x^5 ) + xQ(x^5 ) +x^2R(x^5) ) is divisible by ( x^5-1)

so we have , if a fifth complex root of unity is w then

P(1) + wQ(1) +w^2R(1) =0 ( as w !=1 )..................(1)

That is w is aroot of the eqn

P(1) + xQ(1 ) + x^2R(1 ) =0

similarly w^2, w^3 ,w^4 are also the other fifth complex root of unity and  if we substitute them one by one in the given eqn , it is clear that w^2 , w^3 , w^4 are also distinct roots of the  quadratic eqn .

So , this quadratic eqn has got more than 2 roots.

So , from the corrolary of the fundamental theorem of algebra , it is clear that the eqn is an identity and each of the coefficient of x are zero .

Hence proved .

suppose a|b , then for any non zero m we have , am|bm

does this help u ?

That is nice. Again, this is a problem where knowledge of congruences makes the problem almost an oral one

To see a similar iinstance of this method, have a look at http://www.goiit.com/posts/list/algebra-question-68176.htm#336256

Back to this problem:

Let f(x) = x⁴+x³+x²+x+1

Again we have,

Similar results hold for Q and R.

You can easily extend this to one more degree i.e for the polynomial: