hmm.. let's look at this equation $(mod 4)$ $x^{2}+3x+y-1=0 (mod 4) \\ x_{1,2}=\frac{-3+-\sqrt{13-4y}}{2}(mod 4)$ let's choose $y=1$: $x_{1}=\frac{-3+3}{2}=0 (mod 4)$ so the infinite pairs $(4k,1)$ with the positive integer $k$ are solutions of the equation $(mod 4)$ but not always of the equation...... oh you can choose $y=3$, so that $x_{2}=\frac{-3-1}{2}=-2 (mod 4)$ and $(2,3)$ is a solution for $p=3$, but $3\not|2$...

I don't think so... If $p=3$, then the equation becomes $5x^{2}-4y^{2}+3x+5y=1$ And if $(x,y) = (2,3)$, we find out that $LHS = 5$.. Actually, when I tried to solve this problem, I can't find any single solution.. Even for $p=3$, so strange..

We have $x^{2}=((p+1)(x+y)-1)(y-x-1)$ and $(y-x-1,(p+1)(x+y)-1) \vdots p$

Well, it means that $(p+1)(x+y)-1$ is a square, but if we take $p=3$, then $4(x+y)-1 \equiv 3(mod4)$.. So, $(y-x-1,(p+1)(x+1)-1) > 1$

Proof of the second part:(please let me know if i'm mistaken) consider the equation modulo $p$. we have $2x^{2}-y^{2}+2y-1\equiv 0(\mod p)$ that is $2x^{2}\equiv (y-1)^{2}(\mod p)$ which is possible only if $y-1\equiv 0 (\mod p)$ so $x\equiv 0(\mod p)$.$\blacksquare$

@nayel : Counter Example : $p = 7, y = 3, x = 3$ But of course, it's not a solution...

RDeepMath91 wrote: I don't think so... If $p=3$, then the equation becomes $5x^{2}-4y^{2}+3x+5y=1$ And if $(x,y) = (2,3)$, we find out that $LHS = 5$.. Actually, when I tried to solve this problem, I can't find any single solution.. Even for $p=3$, so strange.. yes, that was my problem: it was a solution for the equation mod 4, but i think i didn't find one for the "normal" equation (but i have to look for my paper, first....)

bumping coz there aren't any solutions..

For $p\mid x_0$, do this. Letting $z=y-1$, we have first through modulo $p$ that $z^2 \equiv 2x^2\pmod{p}$. In particular, our goal is to show, $p\mid x\iff y\equiv 1 \pmod{p}$. Now, note that, $$ x^2 = (z-x)((p+1)(z+x)+p). $$Suppose that $q\mid z-x$ is a prime number. Clearly, $q\mid x$, hence, $q\mid z$. Now, note that, $(p+1)(z+x)+p \equiv p\pmod{q}$. Unless $q\mid z-x$ (in which case, $z\equiv x \pmod{p}$, hence, $2x^2-z^2 \equiv x^2 \equiv 0 \pmod{p}$, thus the claim), we get $(z-x,(p+1)(z+x)+p)=1$. However, this gives a contradiction; since, this means $(p+1)(z+x)+p$ is a perfect square; however in modulo $4$, we have $p+1 \equiv 0 \pmod{4}$, thus, $(p+1)(z+x)+p\equiv 3\pmod{4}$, which cannot be a perfect square. If I find one, I'll post my solution to the infinitude, too.

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Here is a solution of this Diophantine equation with an explicit formula for all solutions. My solution also proves that $p | x_0$ if $(x,y) = (x_0,y_0)$ is a solution of the given equation. We have given a prime $p \equiv 3 \pmod{4}$ and two positive integers $x$ and $y$ satisfying the Diophantine equation $(1) \;\; (p + 2)x^2 - (p + 1)y^2 + px + (p + 2)y = 1$. Equation (1) is equivalent to $(2) \;\; x^2 = (y - x - 1)((p + 1)(x + y) - 1)$. Let $d = GCD(y - x - 1,(p + 1)(x + y) - 1)$. Hence according to (2) there exists two positive coprime integers $u$ and $v$ s.t. $(3) \;\; y - x - 1 = du^2$, $(4) \;\; (p + 1)(x + y) - 1 = dv^2$, $(5) \;\; x = duv$. Now $d | x$, $d | y - x - 1$ and $d | (p + 1)(x + y) - 1$ by (2), yielding $d | y - 1$ and $d | (p + 1)y – 1$. Consequently $d | -(p + 1)(y - 1) + (p + 1)y - 1 1$, i.e. $d | p$, which means $d=1$ or $d=p$. Assume $d=1$. Then $x=uv$ by (5), yielding $y = u^2 + uv + 1$ by (3), which according to (4) implies $(p + 1)(u + v)^2 = v^2 + 1$. We have $4 | p$ since $p \equiv 3 \pmod{4}$, which give us $4 | v^2 + 1$. This contradiction implies $d \neq 1$, which result in $d=p$. Combining (3) and (5) we obtain $(6) \;\; y = pu^2 + puv + 1$. Then by (4)-(6) we obtain $pv^2 + 1 = (p + 1)(x + y) = (p + 1)(pu^2 + 2puv + 1)$, i.e. $(p + 1)u(u + 2v) = v^2 - 1$, or alternatively $(7) \;\; v^2 - 2(p + 1)uv - ((p + 1)u^2 + 1) = 0$. The positive solution of (7) is $(8) \;\; v = (p + 1)u + \sqrt{(p + 1)(p + 2)u^2 + 1}$. Hence there is a positive integer $w$ s.t. $(p + 1)(p + 2)u^2 + 1 = w^2$, which is the Pell’s equation $(9) \;\; w^2 - (p + 1)(p + 2)u^2 = 1$. If $u=1$, then $(2p + 3)^2 < (2w)^2 = (2p + 3)^2 + 3 < (2p + 4)^2$, a contradiction implying $u \neq 1$. By inspection we find that $(u,w) = (2,2p+3) = (u_1,w_1)$ is a solution of (9). This is in fact the fundamental solution of (1), which means all solutions $(u,w) = (u_n,w_n)$ of (1) are given by the (implicitt) formula $(u_n,w_n) = (2p + 3 + 2\sqrt{d})^n$, where $d = (p + 2)(p + 1)$, yielding the (explicit) formulas ${\textstyle (10) \;\; u_n = \frac{1}{2\sqrt{d}}[(2p + 3 + 2\sqrt{d})^n - (2p + 3 - 2\sqrt{d})^n] = \frac{1}{2\sqrt{(p + 2)(p + 1)}}[(\sqrt{p + 2} + \sqrt{p + 1})^{2n} - (\sqrt{p + 2} - \sqrt{p + 1})^{2n}]}$ and ${\textstyle (11) \;\; w_n = \frac{1}{2}[(2p + 3 + 2\sqrt{d})^n + (2p + 3 - 2\sqrt{d})^n] = \frac{1}{2}[(\sqrt{p + 2} + \sqrt{p + 1})^{2n} + (\sqrt{p + 2} - \sqrt{p + 1})^{2n}]}$. By (7) we obtain $(12) \;\; v_n = (p + 1)u_n + w_n$. Hence all solutions of equation (1) are given by the formulas $(x,y) = (pu_nv_n, pu_n^2 + pu_nv_n + 1)$, where $u_n$, $v_n$ and $w_n$ are given by the formulas (10)-(12).