Here's a unified argument based on my comments to Scott's post that doesn't use quadratic reciprocity in any form.  Suppose n=2 and p >= 5, and lift each line of slope i in Y(2,p) to a point (a<sub>i</sub>+pb<sub>i</sub>, ia<sub>i</sub>+pc<sub>i</sub>).

Since each pair of lifts should give a basis of Z<sup>2</sup> and thus a matrix with determinant \pm 1, taking each pair from among i=1,2,k+2 (with 1 <= k <= p-3) gives us conditions

a<sub>1</sub>a<sub>2</sub> = \pm 1 (mod p)

k*a<sub>2</sub>a<sub>k+2</sub> = \pm 1 (mod p)

(k+1)*a<sub>1</sub>a<sub>k+2</sub> = \pm 1 (mod p).

Combining the first two gives ka<sub>2</sub><sup>2</sup>*a<sub>1</sub>a<sub>k+2</sub> = \pm 1, or a<sub>2</sub><sup>2</sup> = \pm(1+1/k) (mod p).

But for k=1 this gives us a<sub>2</sub><sup>2</sup> = \pm 2, and for k=2 we get a<sub>2</sub><sup>2</sup> = \pm (1 + (p+1)/2) = \pm (p+3)/2, so either (p+3)/2 = 2 (mod p) or (p+3)/2 = -2 (mod p).  These imply p=1 and p=7, respectively, so already the only possible solution is p=7.  But if p=7 then k=3 gives a<sub>2</sub><sup>2</sup> = \pm 6, which is not \pm 2 (mod 7), so that doesn't work either.  Thus a lift with n=2 can only possibly exist if p is 2 or 3.