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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_i+pb_i, ia_i+pc_i).

Since each pair of lifts should give a basis of Z² 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

a₁a₂ = \pm 1 (mod p)

k*a₂a_k+2 = \pm 1 (mod p)

(k+1)*a₁a_k+2 = \pm 1 (mod p).

Combining the first two gives ka₂²*a₁a_k+2 = \pm 1, or a₂² = \pm(1+1/k) (mod p).

But for k=1 this gives us a₂² = \pm 2, and for k=2 we get a₂² = \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 a₂² = \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.