$9n^2-4$ only generates one prime? Why?

Question

Instead of doing the work I was supposed to be doing, I played around with some numbers, and I noticed that for $n\in\mathbb{N}$, $9n^2-4$ only seems to generate a prime for $n=1$.

Can anyone explain why?

EDIT: Please don't judge me.

Answer 1

$$9n^2 - 4 = (3n-2)(3n+2)$$

which is composite for $\forall n\gt 1$.

Answer 2

On the other hand, if $n$ is a Markov number, also http://oeis.org/A002559 then $9n^2 - 4$ is the sum of two squares, which you don't see every day.

$$ 9n^2 - 4 = x^2 + y^2 $$

               n        9n^2 - 4             x      y
   1           1              5              2      1
   2           2             32              4      4
   3           5            221             14      5
   4          13           1517             34     19
   5          29           7565             86     13
   6          34          10400            100     20
   7          89          71285            266     23
   8         169         257045            502     71
   9         194         338720            556    172
  10         233         488597            679    166
  11         433        1687397            961    874
  12         610        3348896           1636    820
  13         985        8732021           2786    985
  14        1325       15800621           3875    886
  15        1597       22953677           4286   2141
  16        2897       75533477           7586   4241
  17        4181      157326845          12341   2242
  18        5741      296631725          17197    946
  19        6466      376282400          19372   1004
  20        7561      514518485          22486   2983
  21        9077      741527357          26971   3754
  22       10946     1078334240          32716   2828
  23       14701     1945074605          43709   5882
               n      9n^2 - 4             x      y