Here's a second proof of Claim 1 (I guess the others can be taken care of similarly) that perhaps explains a little bit better what is going on. Assume first that $3x^2 + 1 = py^2$ has an integral solution. Then $\eta = x\sqrt{3} + y \sqrt{p}$ is a unit in ${\mathbb Q}(\sqrt{3},\sqrt{p})$ satisfying $\eta^2 = t + u\sqrt{p}$; thus $\eta^2$ is an odd power of the fundamental unit $\varepsilon$ of ${\mathbb Q}(\sqrt{3p})$. For $p = 829$ we have $\varepsilon = 18982087189657 + 380632678652\sqrt{3p}$, and using the observation that $(2 \cdot 18982087189657 - 2)/3 = 4 \cdot 1778674^2$ (see the proof below) we find $\eta = 1778674 \sqrt{3} + 106999 \sqrt{829}$.

For proving the existence of a solution we simply work backwards (essentially this a classical descent on Pell conics). We start with the fundamental solution $(t, u)$ of $t^2 - 3pu^2 = 1$ and write this equation in the form $(t-1)(t+1) = t^2 - 1 = 3pu^2$. The fact that the fundamental unit has norm $+1$ (the discriminant is divisible by $3$) and that $(t,u)$ is fundamental implies that $3$ and $p$ divide different factors. Using elementary congruences and the fact that $(2/p) = -1$ and $(3/p) = +1$ it is easy to show that the only possibility is $$ t-1 = 6a^2, \quad t+1 = 2pb^2, $$ which implies $1 = pb^2 - 3a^2$.

Here's a second proof of Claim 1 (I guess the others can be taken care of similarly) that perhaps explains a little bit better what is going on. Assume first that $3x^2 + 1 = py^2$ has an integral solution. Then $\eta = x\sqrt{3} + y \sqrt{p}$ is a unit in ${\mathbb Q}(\sqrt{3},\sqrt{p})$ satisfying $\eta^2 = t + u\sqrt{3p}$; thus $\eta^2$ is an odd power of the fundamental unit $\varepsilon$ of ${\mathbb Q}(\sqrt{3p})$. For $p = 829$ we have $\varepsilon = 18982087189657 + 380632678652\sqrt{3p}$, and using the observation that $(2 \cdot 18982087189657 - 2)/3 = 4 \cdot 1778674^2$ (see the proof below) we find $\eta = 1778674 \sqrt{3} + 106999 \sqrt{829}$.

For proving the existence of a solution we simply work backwards (essentially this a classical descent on Pell conics). We start with the fundamental solution $(t, u)$ of $t^2 - 3pu^2 = 1$ and write this equation in the form $(t-1)(t+1) = t^2 - 1 = 3pu^2$. The fact that the fundamental unit has norm $+1$ (the discriminant is divisible by $3$) and that $(t,u)$ is fundamental implies that $3$ and $p$ divide different factors. Using elementary congruences and the fact that $(2/p) = -1$ and $(3/p) = +1$ it is easy to show that the only possibility is $$ t-1 = 6a^2, \quad t+1 = 2pb^2, $$ which implies $1 = pb^2 - 3a^2$.

Here's a second proof of Claim 1 (I guess the others can be taken care of similarly) that perhaps explains a little bit better what is going on. Assume first that $3x^2 + 1 = py^2$ has an integral solution. Then $\eta = x\sqrt{3} + y \sqrt{p}$ is a unit in ${\mathbb Q}(\sqrt{3p})$ satisfying $\eta^2 = t + u\sqrt{p}$; thus $\eta^2$ is an odd power of the fundamental unit $\varepsilon$ of ${\mathbb Q}(\sqrt{3p})$. For $p = 829$ we have $\varepsilon = 18982087189657 + 380632678652\sqrt{3p}$, and using the observation that $(2 \cdot 18982087189657 - 2)/3 = 4 \cdot 1778674^2$ (see the proof below) we find $\eta = 1778674 \sqrt{3} + 106999 \sqrt{829}$.

For proving the existence of a solution we simply work backwards (essentially this a classical descent on Pell conics). We start with the fundamental solution $(t, u)$ of $t^2 - 3pu^2 = 1$ and write this equation in the form $(t-1)(t+1) = t^2 - 1 = 3pu^2$. The fact that the fundamental unit has norm $+1$ (the discriminant is divisible by $3$) and that $(t,u)$ is fundamental implies that $3$ and $p$ divide different factors. Using elementary congruences and the fact that $(2/p) = -1$ and $(3/p) = +1$ it is easy to show that the only possibility is $$ t-1 = 6a^2, \quad t+1 = 2pb^2, $$ which implies $1 = pb^2 - 3a^2$.

Here's a second proof of Claim 1 (I guess the others can be taken care of similarly) that perhaps explains a little bit better what is going on. Assume first that $3x^2 + 1 = py^2$ has an integral solution. Then $\eta = x\sqrt{3} + y \sqrt{p}$ is a unit in ${\mathbb Q}(\sqrt{3},\sqrt{p})$ satisfying $\eta^2 = t + u\sqrt{p}$; thus $\eta^2$ is an odd power of the fundamental unit $\varepsilon$ of ${\mathbb Q}(\sqrt{3p})$. For $p = 829$ we have $\varepsilon = 18982087189657 + 380632678652\sqrt{3p}$, and using the observation that $(2 \cdot 18982087189657 - 2)/3 = 4 \cdot 1778674^2$ (see the proof below) we find $\eta = 1778674 \sqrt{3} + 106999 \sqrt{829}$.

For proving the existence of a solution we simply work backwards (essentially this a classical descent on Pell conics). We start with the fundamental solution $(t, u)$ of $t^2 - 3pu^2 = 1$ and write this equation in the form $(t-1)(t+1) = t^2 - 1 = 3pu^2$. The fact that the fundamental unit has norm $+1$ (the discriminant is divisible by $3$) and that $(t,u)$ is fundamental implies that $3$ and $p$ divide different factors. Using elementary congruences and the fact that $(2/p) = -1$ and $(3/p) = +1$ it is easy to show that the only possibility is $$ t-1 = 6a^2, \quad t+1 = 2pb^2, $$ which implies $1 = pb^2 - 3a^2$.