Solve $an^2+bn+c=k^2$ over the integers.

Question

I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.

$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$

After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535\approx\frac52\sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.

So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?

Answer 1

Complete the square, and you get $$ 5 \left(n + \frac{7}{5}\right)^2 - \frac{44}{5} = k^2 $$ or $$ (5n+7)^2 - 44 = 5 k^2 $$ With $x = 5n+7$, this is a Pell-type equation $$ x^2 - 44 = 5 k^2 $$ where you only want solutions where $x \equiv 2 \mod 5$.