solve in integers $d^4 \pm 1 = 2e^2, e = (pc^2 - 3d^2)/4.$

Question

Solve in integers the diophantine equations $d^4 \pm 1 = 2e^2, e = (pc^2 - 3d^2)/4.$ Assume $p$ is prime.

Clearly $d$ must be odd, and so $d^4 \equiv 0,1\mod 8$.

I'm pretty sure the solution is $d = 1, e =\pm 1, 0, (p,c) = (7, \pm 1), (3, \pm 1)$, but I'm not sure how to show this. I know that in the $-1$ case, $(d^2 - 1)(d^2 + 1)=2e^2$, so $d^2 - 1, d^2 + 1 = u^2, 2v^2$ for some coprime integers $u,v$ such that $e = uv$. Assume $d > 1$. So in particular, since $d^2 - 1$ and $d^2 + 1$ have the same parity, $u$ must be even. So write $u = 2k, v = 2a + 1$. Then $u^2 = 4k^2, 2v^2 = 2(4a^2 + 4a + 1)$. If $u^2 = 2v^2 + 2$, then $4k^2 = 4(2a^2 + 2a + 1)\Rightarrow k^2 = 2a^2 + 2a + 1$. So $k = 2a_2 + 1$ is odd. $(2a_2 + 1)^2 = 2a^2+2a+1\Rightarrow 2a_2(a_2 + 1) =a(a+1).$ The function $x\mapsto x^2 + x$ is clearly strictly increasing. But I'm not sure how to proceed from here.

Clarification: the question was directly taken from the following:

Answer 1

$\ \bullet d^4-1=2e^2$

This curve $v^2=2d^4-2$ has only two integral points, according to the magma online calculator as follows.

IntegralQuarticPoints($[2,0,0,0,-2]$);

It says that all integral points are $( \pm 1 , 0)$.

Hence we get $(d,e)=(\pm 1 , 0)$ then $pc^2=3 \implies (c,p)=(\pm 1 , 3)$.

$\ \bullet d^4+1=2e^2$

Similarly, this curve $v^2=2d^4+2$ has only four integral points, according to the magma online calculator as follows.
IntegralQuarticPoints($[2,0,0,0,2]$);

It says that all integral points are $( \pm 1 , \pm 2)$.

Hence we get $(d,e)=(\pm 1 , \pm 1)$ then $pc^2=7 \implies (c,p)=(\pm 1 , 7)$.

See related info MO

Answer 2

We have:

$d^4-e^2=(d^2-e)(d^2+e)=(e-1)(e+1)$

1):

$e+1=d^2+e\rightarrow d=\pm 1$

$\Rightarrow pc^2-3=4e\rightarrow pc^2\equiv 3\bmod 4 \rightarrow c=\pm1 , p= 7, 11, 13, 17...$

because p is prime.

Corresponding e are:

$e=0,\pm1,2$

2):

$d^2-1=2e^2\rightarrow (d-1)(d+1)=2e^2$

Case 1:

$d+1=2\rightarrow d=1$

$d-1=e^2\rightarrow e=0$

$\rightarrow pc^2=3\rightarrow c=\pm1, p=3$

So we have $(c, d, e, p)=( \pm1, 1, 0, 3) $

Case 2):

$d-1=3\rightarrow d=3$

$d+1=e^2\rightarrow e=\pm 2$

For $e=2$ we have:

$pc^2-3\times 3^2=8\rightarrow c=\pm1, p=35$

For $e=-2$we have:

$pc^2-27=-8\rightarrow pc^2=19$

which gives :

$c=\pm1$ and $ p=19$.

Answer 3

In your argument, when you get $u^2=d^2\pm1$, you have two integer squares differing by 1, so $u$ and $d$ can only be 0 and 1, giving only the solution $d=1$, $e=0$. The other solutions arise from $d^4+1=2e^2$, as explained in this answer: Solve in $\mathbb{Z}$ the equation $x^4 + 1 = 2y^2$.