Integer solutions to the equation $7x^2 = y^2+y+1$

Question

While investigating the related equation $7^n = m^2 + m+1,$ I managed to quite quickly handle the case that $n$ is even. If $n$ is odd, we may let $x = (n-1)/2.$ This now reduces to the question in the title of this post.

However, this is not really a reduction since the equation we obtain is more general. Nontheless, I recognize that solving problems often involves generalizing and solving the general problem from a higher vantage point. This is what I am trying to do here, and I would appreciate if anyone had any ideas on showing that $7x^2 = y^2 + y + 1, x, y > 0$ has only the solutions $(1,2), (7,18)$ (checked up to $y \le 100$).

Answer 1

Let's explain @MichaelRozenberg's strategy. Since $(2y+1)^2-7(2x)^2=-3$, we can find the solutions as follows:

• Find the fundamental solution of $a^2-7b^2=1$: it's $a=8,\,b=3$.
• Do the same thing with $c^2-7d^2=-3$, viz. $c=2,\,d=1$.
• The most general solution of the latter equation is therefore $c+d\sqrt{7}=(2\pm\sqrt{7})(8+3\sqrt{7})^n,\,n\in\Bbb Z$. Now we just need the solutions with $c$ odd, $d$ even (but happily, those conditions are equivalent).

Answer 2

There are infinitely many solutions.

For example, $$x=\frac{(14-5\sqrt7)(127-48\sqrt7)^n+(14+5\sqrt7)(127+48\sqrt7)^n}{28}$$ and $$y=\frac{(5-2\sqrt7)(127-48\sqrt7)^n+(5+2\sqrt7)(127+48\sqrt7)^n-2}{4}$$

For $n=0$ we obtain: $(x,y)=(1,2).$

For $n=1$ we obtain Mason's solution: $(247,653).$

For $n=2$ we obtain: $(62737,165976).$