P+1 can either end up working in a group of order p1 or p+1, depending on whether x_{0}^{2}  4 is a quadratic residue modulo p or not. Since we don't know p in advance, we can't tell which one it was until after p was found. If it was actually p1, the GMPECM prints the message you quoted. The effect of getting either p1 or p+1 is pretty fundamental to how P+1 works, we don't know how to construct a quadratic extension of GF(p) reliably without knowing p. The extra exponentiation etc. we do in GMPECM is not related to this.
Alex
