A Pell equation inside a Pell equation While working on another problem (see https://mathoverflow.net/questions/143599/solving-the-quartic-equation-r4-4r3s-6r2s2-4rs3-s4-1), I found the following equation to be solved:
$$ \bigl((r+s)^2-2s^2\bigr)^2 - 2(2rs)^2 = 1.  \tag{*}$$
Conjecture: There are no solutions with $r > 3$; put another way the solution $(r,s)=(3,2)$ is the only solution with $r>s>0$.
Clearly, $(*)$ is the Pell equation $u^2-2v^2=1$, the fundamental solution of which is [perhaps coincidentally?] $(u,v)=(3,2)$. However, in this case, every other solution $u$ is of the same form, i.e., $u=(r+s)^2-2s^2$. This leads me to believe there is some sort of elementary descent argument that can prove the conjecture. Can anyone point me in the right direction?
Thanks, Kieren.
EDITED to incorporate Will Jagy's observation about "every [other] solution".
EDIT: In addition to (*), the following similar Pell equation is simultaneously true:
$$
    \bigl((r-s)^2-2s^2\bigr)^2 - 2(r^2-s^2)^2 = -1.
$$
EDIT (2014/06/05): Is there a way to prove that these two simultaneous Pell equations force $(r,s)$ to be a solution to $X^2-2Y^2=\pm1$?
 A: [Edited mostly to add the uniqueness of the $(119,120,169)$ triangle
and the elliptic-curve connection]
The conjecture is correct, and equivalent to a famous theorem of Ljunggren
(1942, see references below).  No really simple proof is known,
and it might be that such a proof is not possible; the existence of
several solutions such as $(3,2)$ suggests it's hard to get a proof
with elementary techniques $-$ for instance an attempt to use just
congruences is bound to fail because any congruence condition must
allow the two orbits $\{ (0,\pm1), (1, \pm0) \}$ and
$\{(\pm3,\pm2), (\pm2,\mp3)$ of small solutions.
The left-hand side $\left( (r+s)^2 - 2s^2 \right)^2 - 2(2rs)^2$
is a homogeneous quartic polynomial
$$
Q(r,s) = r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4.
$$
Hence we are dealing with the
Thue equation
$Q(r,s) = 1$.  Thue proved that such an equation has only finitely many
solutions.  His proof is ineffective, i.e. does not yield a method for
provably finding all solutions; but later results do make this possible.
One technique is implemented in several computer algebra packages,
including gp in which the command
thue(thueinit(r^4+4*r^3-6*r^2-4*r+1),1)

yields the expected list
[[-2, 3], [2, -3], [0, 1], [0, -1], [3, 2], [-3, -2], [1, 0], [-1, 0]]

and reports only 14 milliseconds' computing time on my laptop.
But this relies on some quite difficult theorems to bound the search.
The quartic $Q$ is quite nice: it has fourfold symmetry generated by
$(r,s) \mapsto (s,-r)$, and the roots of $Q(r,1)=0$ generate the
cyclic quartic field ${\bf Q}(\cos \pi/8) = \frac12 \sqrt{2+\sqrt 2}$.
But ironically this symmetry makes it hard to use a more elementary
technique, Skolem's $p$-adic method, which often succeeds in showing
that a moderately large list $L$ of solutions is complete by showing
that any other solution must be congruent to one of the solutions in $L$
modulo an arbitrarily large power of $p$.  Ljunggren thus resorted to
applying the method not over ${\bf Q}(\cos \pi/8)$ but over some quadratic
extension, which made his proof even harder to carry out and to follow
(though still not as hard as the bounds exploited by the thue function).
The equation in the title of Ljunggren's paper is not $Q(r,s)=1$ but
$X^2 + 1 = 2Y^4$, which has large integer solutions $(X,Y) = (\pm 239, \pm 13)$
[NB $(1+\sqrt2)^7 = 239 + 13^2 \sqrt 2$] in addition to the clear
$(X,Y) = (\pm 1, \pm 1)$.  Ljungreen showed that there are no other solutions.
Thanks to the identity
$$
Q(r,s)^2 + Q(r,-s)^2 = 2(r^2+s^2)^4
$$
it follows that any solution of $Q(r,s) = 1$ has $r^2+s^2 = 1$ or $13$,
which soon yields the desired result.
Conversely any integer solution of $2X^2 + 1 = Y^4$
yields an integer solution of $Q(r,s) = 1$, as
Steiner and Tzanakis show in Section II of their paper; they use
arithmetic in the Gaussian ring ${\bf Z}[i]$, but one can instead
write the Ljunggren's equation as a primitive Pythagorean triple
$((X-1)/2, (X+1)/2, Y^2)$ and apply the parametrization $(m^2-n^2,2mn,m^2+n^2)$
of Pythagorean triples twice, first to obtain
$$
m^2-n^2-2mn = \pm 1, \quad m^2+n^2 = Y^2
$$
and then to the resulting triple $(m,n,Y)$.  Section III then gives a proof
of the solution of $Q(r,s) = 1$ which the authors claim is
simpler than Ljunggren's, but it requires "a deep result
of Mignotte and Waldschmidt on linear forms in logarithms" (presumably
the same result that the gp program exploits), so the
Steiner-Tzanakis proof ends up even more complicated than Ljunggren's.

Ljunggren, W.:, Zur Theorie der Gleichung $x^2 + 1 = Dy^4$,
  Avh. Norske, Vid. Akad. Oslo 1 #5 (1942)
Steiner, R., and Tzanakis, N.:
  Simplifying the Solution of Ljunggren's Equation $X^2 + 1 = 2Y^4$,
  Journal of Number Theory 37 (1991), 123-132

P.S. 1) The Ljunggren/Pythagorean-triple connection also gives
the following equivalent formulation: There is a unique nearly-isosceles
Pythagorean triangle, namely $(119, 120, 169)$, whose hypotenuse is a square.
(I use "nearly isosceles" to mean $(a,b,c)$ with $a-b = \pm 1$.)
2) If $Q(r,s) = 1$ then $(r,s,1)$ is a integer solution of $Q(r,s) = u^2$.
If we seek rational solutions,
we get an elliptic curve $E$ isomorphic with $y^2 = x^3 - 2x$
[explicitly $u/s^2 = t^2+2t-1 - 2x$ where $t=r/s=(y-x)/(x-2)$].
By standard descent techniques one shows that $E$ has infinitely many points,
each of which is either $kP$ or $kP+T$ where $P$ is the point $(x,y)=(-1,1)$
and $T$ is the 2-torsion point $(x,y)=(0,0)$.  The symmetries of the problem
let us assume without loss of generality that it's $kP$ with $k$ odd.
At this point we can use $p$-adic techniques to show that $k = \pm 1$,
which recovers the known solutions.  For example, if we use $p=3$
then it seems that the numerator of $u/s^2$ is congruent to $1 \bmod 3^f$
iff $k \equiv \pm 1 \bmod 3^{f-1}$, whence $u=1$ happens only
at the known $k = \pm 1$.  A proof along these lines, though still
far from elementary, might satisfy Mordell's desire for a proof
simpler than Ljunggren's, as well as Kieren MacMillan's hope for
a proof that uses descent (though descent is not the only ingredient).
A: The fundamental solution to Pell's equation $x^2 - 2 y^2 = 1$ is $x_0 = 3$, $y_0 = 2$. All other solutions can be expressed by:
$$
x_n - y_n \sqrt{2} = (x_0 - y_0 \sqrt{2})^n
$$
this is:
\begin{align}
x_{n + 1} - y_{n + 1} \sqrt{2}
  &= (x_n - y_n \sqrt{2}) \cdot (3 - 2 \sqrt{2}) \\
  &= (3 x_n + 4 y_n) - (2 x_n + 3 y_n) \sqrt{2}
\end{align}
Mangling, spindling, and otherwise manhandling the above gives the recurrences:
\begin{align}
x_{n + 1} &= 3 x_n + 4 y_n \\
y_{n + 1} &= 2 x_n + 3 y_n
\end{align}
Now you have:
\begin{align}
x_n
  &= (r + s)^2 - 2 s^2 \\
y_n
  &= 2 r s
\end{align}
so that:
$$
r^2 - s^2 = x_n - y_n
$$
One solution to this is given by $r - s = 1$ so that $r^2 - s^2 = 2 s + 1$,
as long as $x_n - y_n$ is odd you get a solution:
$$
r = \frac{x_n - y_n + 1}{2}, s = \frac{x_n - y_n - 1}{2}
$$
It turns out that both $x_n$ and $y_n$ satisfy the recurrence:
$$
u_{n + 2} = 6 u_{n + 1} - u_n
$$
so that $x_n - y_n$ also satisfies this recurrence. Initial values are $x_0 - y_0 = 1$, $x_1 - y_1 = 5$, and by induction through the recurrence $x_n - y_n$ is always odd. 
A: For Pell's equation: $x^2-2y^2=1$ following decision $(x_1;y_1)$ You can find previous knowledge $(x;y)$ formula:
$x_1=3x\pm4y$
$y_1=2x\pm3y$
We get a system. In this case we can set any shape these decisions - as we want. Well, it will be necessary to solve this system.Although you can not solve. You can see that at each step on the right is a polynomial of n-th degree, and left the second degree. In integers no solutions. Rather if there is one.
