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Two number sequences defined as: $$a_1=b_1=1,$$ $$a_k=3a_{k-1}+4b_{k-1},\quad b_k=2a_{k-1}+3b_{k-1}\quad \forall 2\le k\in\mathbb{Z}^+$$

Is it possible to find all $k$s that make $a_k$ perfect squares?

update

It is possible to find that:

$$a_n=\dfrac{1}{2} \left(1+\sqrt{2}\right) \left(\left(3+2 \sqrt{2}\right)^{n-1}-\left(3-2 \sqrt{2}\right)^n\right),\quad n\in\mathbb{Z}^+$$

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    $\begingroup$ Have you found any examples (other than $k=1$)? $\endgroup$ – lulu Jun 20 '17 at 20:15
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    $\begingroup$ Hint : first note that $a_k^2-2b_k^2=-1$... $\endgroup$ – Mike Bennett Jun 20 '17 at 20:20
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    $\begingroup$ @Mike Bennett What is the rest? $\endgroup$ – Michael Rozenberg Jun 20 '17 at 20:40
  • $\begingroup$ $a_{n+2}=6a_{n+1}-a_n$, which gives $a_n$, which is very ugly. $\endgroup$ – Michael Rozenberg Jun 20 '17 at 20:52
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    $\begingroup$ @MichaelRozenberg it is also true that $b_{n+2} = 6 b_{n+1} - b_n.$ This is Cayley-Hamilton for the given 2 by 2 coefficient matrix. $\endgroup$ – Will Jagy Jun 20 '17 at 21:14
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Just to fill in the early parts, the given coefficient matrix is an "automorphism" of the quadratic form $x^2 - 2 y^2.$ That is, $$ (3x+4y)^2 - 2 (2x+3y)^2 = x^2 - 2 y^2. $$ So, starting with $a_1^2 - 2 b_1^2,$ we will always get the same value for $a_j^2 - 2 b_j^2.$

Next, the trace is $6$ and determinant $1,$ if we call the matrix $$ A = \left( \begin{array}{cc} 3 & 4 \\ 2 & 3 \end{array} \right) $$ we get, by Cayley-Hamilton. $$ A^2 - 6 A + I = 0 $$ We also get $$ a_{n+2} = 6 a_{n+1} - a_n, $$ $$ b_{n+2} = 6 b_{n+1} - b_n. $$

The end, which i may not be able to supply, is usually unique factorization, in this case in the ring of integers of $\mathbb Q [\sqrt 2].$ i will see if it is in Mordell's book.

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jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:  
    3   4
    2   3
  Automorphism backwards:  
    3   -4
    -2   3

  3^2 - 2 2^2 = 1

 u^2 - 2 v^2 = -1

Tue Jun 20 15:15:12 PDT 2017

u:  1  v:  1 ratio: 1  SEED   KEEP +- 
u:  7  v:  5 ratio: 1.4
u:  41  v:  29 ratio: 1.41379
u:  239  v:  169 ratio: 1.4142
u:  1393  v:  985 ratio: 1.41421
u:  8119  v:  5741 ratio: 1.41421
u:  47321  v:  33461 ratio: 1.41421
u:  275807  v:  195025 ratio: 1.41421
u:  1607521  v:  1136689 ratio: 1.41421
u:  9369319  v:  6625109 ratio: 1.41421
u:  54608393  v:  38613965 ratio: 1.41421

Tue Jun 20 15:17:12 PDT 2017

 u^2 - 2 v^2 = -1

jagy@phobeusjunior:~$ 

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  • $\begingroup$ I can find that $a_n=\frac{1}{2} \left(1+\sqrt{2}\right) \left(\left(3+2 \sqrt{2}\right)^{n-1}-\left(3-2 \sqrt{2}\right)^n\right)$,but don't know how to do the next step. $\endgroup$ – LCFactorization Jun 20 '17 at 21:55
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    $\begingroup$ @LCFactorization it would appear that Mordell's book says that there is just one solution when BOTH $a,b$ are squares. Even then, he does not write it all out. $\endgroup$ – Will Jagy Jun 20 '17 at 22:13
  • $\begingroup$ May I have the book name ? $\endgroup$ – LCFactorization Jun 21 '17 at 0:44
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    $\begingroup$ @lc Diophantine Equations (1969). $\endgroup$ – Will Jagy Jun 21 '17 at 0:50
  • $\begingroup$ So finding the square values of the $b_k$ amounts to an old problem, solved by Ljunggren in the 1940s. There is no known "easy" solution, but the only square values are $1$ and $169$. The squares in the $a_k$ sequence (or, equivalently, the equation $x^4-2y^2=-1$) are much easier to find -- see Mordell, page 18. $a_k=1$ is the only one. $\endgroup$ – Mike Bennett Jun 21 '17 at 12:25

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