For which integers $q \ge p\ge 1$ with $q^2-2p^2=2$ is $2p^2+1 \pm pq$ an integer square?

Question

The title says it all… I’m looking to prove (in an elementary way, if possible) the following question:

Conjecture: If $q$ and $p$ are positive integers such that $q^2-2p^2=2$ and $2p^2+1 \pm pq$ is an integer square, then $(q,p)=(2,1)$ or $(q,p)=(10,7)$.

I’ve verified the conjecture for $1 \le p \le q \le 6000$ [and counting].

EDIT: Notice that $(q-p)(q+p)=p^2+2$, and the first solution has $q+p=3$ [with $q-p=1$], and the second solution has $q-p=3$ [with $q+p=17$]. Not sure if that’s helpful, but I thought it an interesting connection…

Answer 1

For some integer $r$, we have

$$2p^2+1\pm pq=r^2,$$

which is to say

$$ p^2 q^2=(r^2-2p^2-1)^2.$$

Since $q^2-2p^2=2$, then

$$p^2 (2p^2+2)=(r^2-2p^2-1)^2.$$

Rearranging this as a quadratic in $p^2$ yields

$$2p^4-(4r^2-2)p^2+(r^2-1)^2=0.$$

The discriminant must be a perfect square, and thus $$\Delta=2r^4-1=s^2$$ for some integer $s$. Hence [by Ljunggren’s theorem] the only solutions are $$(r, s)=(\pm1, \pm1), (\pm13, \pm 239).$$

Answer 2

Just for speed, you need consider only your two initial pairs and then $$ q_{n+2} = 6 q_{n+1} - q_n, $$ $$ p_{n+2} = 6 p_{n+1} - p_n. $$

x:  2  y:  1 ratio: 2  SEED   KEEP +- 
x:  10  y:  7 ratio: 1.42857
x:  58  y:  41 ratio: 1.41463
x:  338  y:  239 ratio: 1.41423
x:  1970  y:  1393 ratio: 1.41421
x:  11482  y:  8119 ratio: 1.41421
x:  66922  y:  47321 ratio: 1.41421
x:  390050  y:  275807 ratio: 1.41421
x:  2273378  y:  1607521 ratio: 1.41421
x:  13250218  y:  9369319 ratio: 1.41421