Find the Different Integer Solution to the Equation 5x
EDIT, January 2015: Conway's little book is available at http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf
I also put four related excerpts, all with the prefix indefinite_binary, at OTHER. Dmitry says the computer zakuski is being decommissioned, hope it continues to work through late January. I especially like Stillwell's presentation. Put it all together, for a Pell form, indeed any form $a x^2 + b xy + c y^2$ with $a > 0, \; b \geq 0, \; c < 0,$ but $b^2 - 4ac$ not a perfect square, we get a diagram that shows all of Conway's information, along with the $(x,y)$ pairs as column vectors, with an explicit illustration of the (proper) automorphism group generator, that being the mapping $(x,y) \mapsto (9x+20y,4x+9 y). $
Did not notice this one ten days ago. There is an explicit structure for representing a number by an indefinite quadratic form. This is chapter one in Conway's The Sensual Quadratic Form. I wrote a little program recently, and no longer make simple arithmetic mistakes in these.
It turns out that all occurrences of $\pm 4$ happen along the "river" for $x^2 - 5 y^2. $
Given any solution to $x^2 - 5 y^2 = \pm 4,$ we gat the same value by switching $(x,y)$ to $$ (9x+20y,4x+9 y). $$ The two by two matrix causing this transformation (on column vectors) is $$ A \; = \; \left( \begin{array}{rr} 9 & 20 \\ 4 & 9 \end{array} \right) , $$ which you can see towards the right of the diagram as the coordinates of the final $1$ and then the final $-5,$ placed side by side. The big theorem is that the entire diagram is periodic. I find the finite set of representatives within one cycle, apply the transformation I wrote arbitrarily many times, and i get all. As there is no $xy$ term in $x^2 - 5 y^2,$ there is a simple $\pm$ symmetry as well.
So, all solutions to $x^2 - 5 y^2 = \pm 4 $ are:
Imprimitive:
+4: $$(2,0), (18,8), (322,144), (5778,2584), (103682,46368), (1860498,832040),\ldots, $$
-4: $$(-4,2), (4,2), (76,34), (1364,610), (24476,10946), (439204,196418),\ldots, $$
Primitive:
+4: $$(3,-1), (7,3), (123,55), (2207,987), (39603,17711), (710647,317811), \ldots, $$
+4: $$(3,1), (47,21), (843,377), (15127,6765), (271443,121393), \ldots, $$
-4: $$(-1,1), (11,5), (199,89), (3571,1597), (64079,28657), (1149851,514229), \ldots, $$
-4: $$(1,1), (29,13), (521,233), (9349,4181), (167761,75025), \ldots, $$
For any position in these sequences, there is a degree two recursion given by
$$ a_{n+2} = 18 a_{n+1} - a_n. $$ For example, $18 \cdot 29 - 1 = 521,$ then $18 \cdot 521 - 29 = 9349. $
Let's see, 3:21 pm. Both Fibonacci and Lucas do the same thing (by six positions), as $$ F_{n+12} = 18 F_{n+6} - F_n, $$ $$ L_{n+12} = 18 L_{n+6} - L_n. $$ So, if the six orbits above satisfy the desired Fibonacci/Lucas conditions, that is a complete proof. If so, one could, carefully, interleave the six orbits in numerical order, perhaps using only the ones with strictly positive entries. See whether that works:
$$ (1,1),(3,1),(4,2),(7,3),(11,5), (18,8),$$ $$ (29,13),(47,21),(76,34),(123,55),(199,89), (322,144),$$ $$(521,233),(843,377),(1364,610),(2207,987),(3571,1597),(5778,2584), $$ $$(9349,4181),(15127,6765),(24476,10946),(39603,17711),(64079,28657),(103682,46368), $$ $$ (167761,75025),(271443,121393),(439204,196418),(710647,317811),(1149851,514229),(1860498,832040), $$ Yep. The only miss is $(2,0),$ as $2$ is not a Lucas number. CORRECTION, FEB. 2015: as is commented elsewhere, it appears fairly common for people to define Lucas number $L_0 = 2,$ http://en.wikipedia.org/wiki/Lucas_number
Ummm; as you can see, $(x,y)$ and $(x,-y)$ may be distinct as far as the orbits, the six lists i wrote.
There is plenty more that could be said; anyway, these give all solutions. Oh, the other business, the "climbing lemma," says that values only increase (in absolute value) when leaving the river. The next layers of values are $\pm 11$ at the continuation of each edge with a light blue $6,$ and $\pm 19$ at the continuation of each edge with a light blue $10.$ So we have done enough to catch all $\pm 4$ already.
Find the Different Integer Solution to the Equation 5x
Source: https://math.stackexchange.com/questions/742181/find-all-integer-solutions-for-the-equation-5x2-y2-4