Let p be an odd prime. The list of p^{th} powers mod θ does not
contain two consecutive nonzero (mod θ) integers if and only if
x^{p} + y^{p} + z^{p} = 0 mod θ implies x = 0 mod
θ or y = 0 mod θ or z = 0 mod θ.

Proof

Suppose x, y, and z are all nonzero mod θ and
x^{p} + y^{p} + z^{p} = 0 mod θ. Then
there is a nonzero (mod θ) integer a such that xa = 1 mod θ, and nonzero integers b
and c with yb = 1 mod θ and zc = 1 mod θ. Then

x^{p} + y^{p} + z^{p} = 0 mod θ
=> (ax)^{p} + (ay)^{p} + (az)^{p} = 0 mod θ
=> (ay)^{p} + (az)^{p} = -1 mod θ

Let k = (ay)^{p} mod θ. Note that k ≠ 0 mod θ since otherwise

(ay)^{p} = 0 mod θ
=> 0 = a^{p}y^{p}b^{p}
= a^{p}(yb)^{p} = a^{p} mod θ

which is impossible if a is nonzero mod θ.

Therefore (θ-az)^{p} = -(az)^{p} = k+1 mod θ. Thus
(ay)^{p} and (θ-az)^{p} are consecutive nonzero
p^{th} powers.

Conversely, if a^{p} = k and b^{p} = k+1 are two
consecutive nonzero (mod θ) integers, then let x=θ-a, y=b, and z=θ-1. Then