When p=3 there is no such x so what about other primes greater than 3

Can you motivate the problem? The condition on k does not look very natural...

The condition on k is essential to guarantee that 1+x+...+x^k permutes the elements of Zp

There is no such x when p=5 so what about p ≥7

The polynomial you consider has no solution mod p.

Using the hypothesis k=1 mod p, You can write it in Z/pZ[X] :

1+2x+...+k.x^(k-1)=\sum_{i=0}^{(k-1)/p-1} x^(ip)\sum_{j=0}^{p-1}(j+1)x^j+x^k

so:

\Phi_p'(x).(x^(k-1)-1)/(x^p-1)+x^k

you can notice that the polynomial (x^(k-1)-1)/(x^p-1) is identically nul for x in Z/pZ different from 0 or 1 and for such values of x x^k is not null. Furthermore, 0 is trivialy not a root of your polynomial and the value at 1 is :

k(k+1)/2=1 mod p with your hypothesis.

If (x, p) = 1 and x isn' t 1 mod p, then we obtain

(x^k - 1)/(x - 1) + x(x^(k-1)-1)/(x-1) + .... + x^(k-1) (x - 1)/(x - 1) =

(x - 1)^-1 [ kx^k - (x^k - 1)/(x - 1)] = 0 (mod p) by Fermat's little Theorem.

Then x = p - 1 satisfies the condition above.