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The Pell equation $(2x+y)^2-5y^2=4$ has fundamental solution $(1,1)$. I need to find three more solutions, I know tha they are the Fibonacci numbers in ordered pairs $(2,3), (5,8),...$, but I don't know how to find them formally. Normally we would compute $x_i=(1+1*\sqrt5)^i$ when the equation the Pell equation is equal to $1$, but this is a little different, it is general (equal to $4$ and there is a $(2x+y)$ term).

Any solutions/hints are greatly appreciated.

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    $\begingroup$ You can use solutions of the equation $a^2-5b^2=1$, like $(a,b)=(9,4)$. Note that when you multiply your equation by this one, the right side is $4$ and the left side can be written as $A^2-5B^2$ for $A,B$ being linear combinations of $2x+y$ and $y$ with coefficients that depend on $a,b$. See Brahmagupta's identity. $\endgroup$
    – plop
    Commented Dec 2, 2021 at 12:48
  • $\begingroup$ What do you mean, you don't know how to find them formally? Are you trying to prove that Fibonacci pairs give solutions to the equation? (If all you need is to find three more solutions and you know, even without proof, that Fibonacci pairs do the trick, then all you have to do is check that $(x,y)=(2,3)$, $(5,8)$, and $(13,21)$ satisfy the equation.) $\endgroup$
    – Barry Cipra
    Commented Dec 2, 2021 at 13:10
  • $\begingroup$ I understand OP to mean that he does not want to use that he knows the answer involves Fib numbers, or failing that, OP would like a natural way to guess that the Fib numbers are involved (at which point, Barry's verification step would work) $\endgroup$
    – Calvin Khor
    Commented Dec 2, 2021 at 13:26
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    $\begingroup$ For $p^2-5q^2=4$ all pair $(p,q)$ get from $p+qz\equiv2((z-1)/2)^j\pmod{z^2-5}$ where $j$=0,2,4,6,... Then $y=abs(q)$ and $x=(p-y)/2$. Finally $(x,y)$=(1,0), (1,1), (2,3), (5,8), (13,21), (34,55), (89,144), (233,377), (610,987),.. $\endgroup$
    – Dmitry Ezhov
    Commented Dec 2, 2021 at 13:49
  • $\begingroup$ @ddswsd FYI, this answer states "... $5q^2 + 4 = n^2$ whose solutions depend on Fibonacci numbers of the form $F_{2m}$". $\endgroup$
    – John Omielan
    Commented Dec 2, 2021 at 21:20

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Expanding out $(2x+y)^2-5y^2=4$ yields $$x^2+xy-y^2=1\tag{1}$$

It is indeed true that the Fibonacci numbers satisfy eq. $(1)$, for example the pair $(1,1)=(F_1,F_2)$. The next solution should then be $(F_3,F_4)=(2,3)$ which works, since $4+10-9=1$.

If $(x,y)=(F_{2n-1}, F_{2n})$ works, then $(F_{2n+1},F_{2n+2})=(x+y,x+2y)$ should work aswell. This can be proven easily as

$$(x+y)^2+(x+y)(x+2y)-(x+2y)^2=x^2+xy-y^2=1$$

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