This specific quadratic came up as part of a puzzle, but the context isn't really important. I just need to find all positive integers $n$, where $5n^2+14n+1$ is a perfect square.
Unfortunately I'm not really a number theorist and I don't know enough "tricks" to make this work out. The only tricks I know are to either (a) recognize this as a Pell equation variant, or (b) represent the quadratic as the sum of two perfect squares, and somehow use Euclid's theorem on Pythagorean triples.
I don't think this is a Pell equation variant, or if it is, I don't see how. When you complete the square you get $5(n+\frac{7}{5})^2 - \frac{44}{5}$ which doesn't seem helpful as the inside of the square isn't an integer.
Similarly I don't see how to see it as a sum of two squares, as 5 is only the sum of two squares in one way -- $1^2+2^2$ -- and then the two squares would have to be of the form $(2n+a)^2 + (n+b)^2$, and so $a^2+b^2=1$, and either $(a,b)=(1,0)$ or $(a,b)=(0,1)$, neither of which work.
So I'm a bit at a loss, and would appreciate any kind of hint.
If you wish to understand this, for example why there are six orbits, i recommend this new book, Weissman, An Illustrated Theory of Numbers, which tells how to draw the Conway topograph.
Your Pell variant comes from $$ (5n+7)^2 - 5 m^2 = 44. $$
My program calls it $w^2 - 5 v^2 = 44.$
There are several orbits for this. You want the ones where $w \equiv 2 \pmod 5,$ so they end in a $2$ or a $7.$ There are infinitely many, it will take some time to describe a recurrence. Anyway, given such a $w,$ then take $$ n = \frac{w - 7}{5} \; \; . $$
Alright, there are six distinct orbits of successful $w.$ In each case, we get a linear recurrence from
$$ \color{blue}{ w_{k+2} = 322 w_{k+1} - w_k} $$ $$ (A).....7, \; 1487, \; 478807, \; 154174367,... $$ $$ (B).....17, \; 5257, \; 1692737, \; 545056057,... $$ $$ (C).....32, \; 10192, \; 3281792, \; 1056726832,... $$ $$ (D).....112, \; 36032, \; 11602192, \; 3735869792,... $$ $$ (E).....217, \; 69857, \; 22493737, \; 7242913457,... $$ $$ (F).....767, \; 246967, \; 79522607, \; 25606032487,... $$
The matching sequences of $n_k$ satisfy
$$ \color{blue}{ n_{k+2} = 322 n_{k+1} - n_k + 448} $$ For example, the first orbit of $n$ actually contains $0,$ not positive quite yet: $$ (A).....0, \; 296, \; 95760, \; 30834872,... $$ $$ (B).....2, \; 1050, \; 338546, \; 109011210,... $$ $$ (C).....5, \; 2037, \; 656357, \; 211345365,... $$ $$ (D).....21, \; 7205, \; 2320437, \; 747173957,... $$ $$ (E).....42, \; 13970, \; 4498746, \; 1448582690,... $$ $$ (F).....152, \; 49392, \; 15904520, \; 5121206496,... $$
I ran a separate thing to just report the first 24 values of $n,$ in order rather than six families:
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