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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.

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    $\begingroup$ Brute forcing the first few values: $1,4,49,676,\dots$ and searching oeis gives this result. $\endgroup$ – JMoravitz Jul 10 '18 at 0:17
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    $\begingroup$ @fleablood Bears what out? The recursion $a(n) = 14 a(n-1)-a(n-2)-6$ shows this is an infinite sequence. $\endgroup$ – Robert Israel Jul 10 '18 at 0:53
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    $\begingroup$ Indeed, writing this as $b^2-3a^2=1$ immediately places it as a Pell equation. $\endgroup$ – Steven Stadnicki Jul 10 '18 at 0:54
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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 $$x_n+y_n\sqrt3=(2+\sqrt3)^n$$

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  • $\begingroup$ Consider adding the recurrence for $(x_n,y_n)$. $\endgroup$ – lhf Jul 10 '18 at 0:51
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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
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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*}

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