# 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.

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

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 $$x_n+y_n\sqrt3=(2+\sqrt3)^n$$
• Consider adding the recurrence for $(x_n,y_n)$. – lhf Jul 10 '18 at 0:51
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:~$./Pell_Target_Fundamental Automorphism matrix: 2 3 1 2 Automorphism backwards: 2 -3 -1 2 2^2 - 3 1^2 = 1 w^2 - 3 v^2 = 1 = 1 Mon Jul 9 17:37:19 PDT 2018 w: 1 v: 0 SEED KEEP +- w: 2 v: 1 SEED BACK ONE STEP 1 , 0 w: 7 v: 4 w: 26 v: 15 w: 97 v: 56 w: 362 v: 209 w: 1351 v: 780 w: 5042 v: 2911 w: 18817 v: 10864 w: 70226 v: 40545 w: 262087 v: 151316 w: 978122 v: 564719 w: 3650401 v: 2107560 w: 13623482 v: 7865521 w: 50843527 v: 29354524  Define sequences$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}\$.