If I understand this correctly, then for p > 5 the vectors (a,0) and (0,a) in Y(2,p) should have to lift to vectors (a+kp,lp) and (mp,a+np) in X(2) which are a basis of Z^2. But this shouldn't be possible unless they form a matrix with determinant \pm 1, i.e. (a+kp)(a+np) - nmp^2 = \pm 1, In particular this requires a^2 = \pm 1 (mod p), but this is impossible for a=2.

This should also generalize in the obvious way to n > 2: we get a^n = \pm 1 (mod p), and if a is a primitive root modulo p then a^n = \pm 1 (mod p) implies a^{2n} = 1 (mod p), which can't be solved for p > 2n+1 (as well as lots of smaller p).

This doesn't eliminate all the cases that don't work, but we already see that for fixed n only finitely many p will work.

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.

If I understand this correctly, then for p > 5 the vectors (a,0) and (0,a) in Y(2,p) should have to lift to vectors (a+kp,lp) and (mp,a+np) in X(2) which are a basis of Z^2. But this shouldn't be possible unless they form a matrix with determinant \pm 1, i.e. (a+kp)(a+np) - nmp^2 = \pm 1, In particular this requires a^2 = \pm 1 (mod p), but this is impossible for a=2.

This should also generalize in the obvious way to n > 2: we get a^n = \pm 1 (mod p), and if a is a primitive root modulo p then a^n = \pm 1 (mod p) implies a^{2n} = 1 (mod p), which can't be solved for p > 2n (as well as lots of smaller p).

This doesn't eliminate all the cases that don't work, but we already see that for fixed n only finitely many p will work.

If I understand this correctly, then for p > 5 the vectors (a,0) and (0,a) in Y(2,p) should have to lift to vectors (a+kp,lp) and (mp,a+np) in X(2) which are a basis of Z^2. But this shouldn't be possible unless they form a matrix with determinant \pm 1, i.e. (a+kp)(a+np) - nmp^2 = \pm 1, In particular this requires a^2 = \pm 1 (mod p), but this is impossible for a=2.

This should also generalize in the obvious way to n > 2: we get a^n = \pm 1 (mod p), and if a is a primitive root modulo p then a^n = \pm 1 (mod p) implies a^{2n} = 1 (mod p), which can't be solved for p > 2n+1 (as well as lots of smaller p).

This doesn't eliminate all the cases that don't work, but we already see that for fixed n only finitely many p will work.