The Diophantine equation
(1) y^2 = px^2 + 1
has certain solutions in x,y,p.
Case 1. For p = 4k+1, x is even and y is odd for all solutions x,y,p.
Proof: Assume x is odd = 2m+1 for some m. Then y is even = 2k so that (1) becomes
(2) 4k^2 = (4k+1)(2m+1)^2 +1
This translates to 4w = 4z+2 for some w,z. Dividing out the 2 we get 2w = 2z+1 which is impossible for all integers w and z.
Therefore x is even and y is odd in (1) for p = 4k+1.
Case 2. For p = 4k+3 if x is odd and y is even for all solutions x,y,p.
If x is even, we have
(3) 4n^2+4n+1 = 4pm^2 +1 = n^2 + n = pm^2. This implies m divides n. We have n = zm for some z coprime with m (uniqueness of factorization). Then (3) becomes
(4) z^2m^2 + zm = pm^2
dividing out m we get z^2m + z = pm. This implies m divides z contrary to z coprime with m. Hence x cannot be even x must be odd.
We verify this by letting x = 2m+1 and y = 2k and substututing this in (1) and multiplying out we have
(5) y^2 = px^2 + 1 = 4k^2 = p(2m+1)^2 + 1. This can be rewritten as
(6) (2k+1)(2k-1) = p(2m+1)^2.
So 2m+1 divides 2k+1 or 2k-1.
Then p divides y-1 or y+1 in (5).
Conjecture: p does not divide x for all solutions of (1).
