Finishing (px- 1) (2x-p) = 2pq 2-(#)

∴p|(px- 1)(2x-p), and (p, px- 1)= 1.

∴p|2x-p, p|x, let x=kp(k∈Z*).

Substitute (#), (KP 2- 1) (2k- 1) = 2q 2.

∫2k- 1 is an odd number, ∴ 2k- 1 | q 2, ∴2k- 1= 1,q or q2.

(1) If 2k- 1= 1, k= 1, x=p, p 2-2q 2 = 1.

Obviously P is an odd number, ∴ p 2 ≡ 1 (mod 8), ∴ 8 | 2q 2, ∴ 2 | q.

∴q=2,p=3,x0=3,y0=4

(2) if 2k-1= q 2, KP 2 = 3, but KP 2 >; =p^2>； =9, contradiction!

(3) If 2k- 1=q, KP 2- 1 = 2Q, the two expressions are subtracted to get Q = K (P 2-2).

∵p^2-2>； 1, ∴ only k= 1, ∴q= 1, contradiction!

To sum up, p=3.