Are there infinitely many positive integer squares of the form $3a^2+1$? Are there infinitely many positive integer squares of the form $3a^2+1$? So I know that it is a square for $a=4,15$. Is there a way to see that there exist infinitely such $a$'s?
I am trying to solve an Olympiad question, and if this is tree, the question will be solved. 
 A: HINT.-The fundamental unit of $\mathbb Q(\sqrt3)$ is $u=2+\sqrt3$ so the solutions $(x_n,y_n)$ of $b^2-3a^2=1$ is given by  $$x_n+y_n\sqrt3=(2+\sqrt3)^n$$
A: The values for $a$ in $b^2 - 3 a^2 = 1$   are
$$ 0, 1, 4, 15, 56,  $$
and obey
$$  a_{n+2} = 4 a_{n+1} - a_n  $$
See the values for v below
The part about the automorphism matrix say that, given a solution $(b,a),$ you get the next pair from
$$ (b,a) \mapsto (2b+3a, b + 2 a )  $$ suach as
$(7,4) \mapsto (26, 15)$
jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:  
    2   3
    1   2
  Automorphism backwards:  
    2   -3
    -1   2

  2^2 - 3 1^2 = 1

 w^2 - 3 v^2 = 1 =   1 

Mon Jul  9 17:37:19 PDT 2018

w:  1  v:  0  SEED   KEEP +- 
w:  2  v:  1  SEED   BACK ONE STEP  1 ,  0
w:  7  v:  4
w:  26  v:  15
w:  97  v:  56
w:  362  v:  209
w:  1351  v:  780
w:  5042  v:  2911
w:  18817  v:  10864
w:  70226  v:  40545
w:  262087  v:  151316
w:  978122  v:  564719
w:  3650401  v:  2107560
w:  13623482  v:  7865521
w:  50843527  v:  29354524

A: Define sequences $a_1=1,a_2=4$ and $a_n=4a_{n-1}-a_{n-2}$ and 
$b_1=2,b_2=7$ and $b_n=4b_{n-1}-b_{n-2}$.
Now show by strong induction that
\begin{eqnarray*}
3a_na_{n-1}+2&=&b_nb_{n-1} \\
3a_n^2+1&=&b_n^2.
\end{eqnarray*}
