Proof:  As p and n are both introspective for (x+a), we have

We must  show h ≡ 0(mod x^r−1).Because (r,p) =1, x^r−1 factors into distinct irreducible h_i(x) over Z_p. Using the Chinese Remainder theorem , we get

 

As x^r−1 divides  h^p, each of the irreducible factors h_i(x) divide h. Hence x^r−1 divides h. Hence the proof.

It is easy to see the introspective numbers are closed under multiplication and that the set of functions for which a given integer is introspective is closed under multiplication.

We can now state a fact as a consequence of the above results.
Every element if the set   is introspective for every polynomial in the
set  . We now define two groups based on these sets that will play a crucial role in the proof.

1. This is a subgroup of Z^*_r since (n,r) =(p,r)=1. Let G be this group and |G|=t.G is generated by n and p modulo r and since  O_r(n) >  log² (n), t > log² (n).
2. Let Q_r(X) be r^th cyclotomic polynomial over F_p . Polynomial Q_r(X) divides X^r−1 and factors into irreducible factors of degree o_r(p) . Let h(X) be one such irreducible factor. Since  o_r(p) >1, the degree of h(X) is greater than one. The second group is the set of all residues of polynomials in P modulo h(X) and p. Let G be this group. This group is generated by elements  X, X+1,X+2,…, X+l in the field F = F_pX/ (h(X)) and is a subgroup of the multiplicative group of F.