hsbhatt sir plz gave me full solution of that question.

(1) if p(x) = 1+x+x²+x³...............x^m

and p(xⁿ) is divisable by  p(x) then find he relation between m and n.

P(x) =

For  P()  to be divisible by P(x), every divisor of  P(x) must be a divisor of P().

The divisors  of  P(x)  are the non-real roots of 

So now if  is a root of  the above  equation  then (x-)  is a divisor of P(x) and must  be a  divisor of  P()

Two  cases  arise:

1. nm+1

Suppose  m+1 is a multiple of n then   and hence P(x) is not a divisor. If  n is prime to m+1 then

and so P(x) is a divisor

2. n>m+1

If n is a multiple of m+1 then   and P(x) is not a divisor

If n r mod (m+1) then where r<m+1 and we can use the same argument as in case 1

In general we can say n r mod (m+1) where gcd(r,m+1) = 1

I am unable to edit my earlier post, but I just want to say that when we say that nr mod m+1 where r is prime to  m+1 is the same as stating that n is prime to m+1.

So the condition is that n and m+1 are relatively prime