Solving s^2-np^2=1

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Solving s^2-np^2=1

Why s²-np²=1 has infinitely many solutions for all 'n'

Let's have a look at the square root of 13, governed by the factor 1298

...	842401	-3037320	-2194919	-1352518	-510117	332284	-177833	154451	-23382	60923	37541	14159	-9223	4936	-4287	649	-2340	-1691	-1042	-393	256	-137	119	-18	47	29	11	-7	4	-3	1	0	1	2	3	4	7	11	18	83	101	119	137	256	393	649	2340	2989	3638	4287	4936	9223	14159	23382	107687	131069	154451	177833	332284	510117	842401	3037320	...
...	-233640	842401	608761	375121	141481	-92159	49322	-42837	6485	-16897	-10412	-3927	2558	-1369	1189	-180	649	469	289	109	-71	38	-33	5	-13	-8	-3	2	-1	1	0	1	1	1	1	1	2	3	5	23	28	33	38	71	109	180	649	829	1009	1189	1369	2558	3927	6485	29867	36352	42837	49322	92159	141481	233640	842401	...
-31	-30	-29	-28	-27	-26	-25	-24	-23	-22	-21	-20	-19	-18	-17	-16	-15	-14	-13	-12	-11	-10	-9	-8	-7	-6	-5	-4	-3	-2	-1	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	26	27	28	29	30	31	32

If we take the set of series apart we find 15 different series. The section indexed 0 to 14 inclusive, is called the 'primary section' of the set of series (as opposed to the secondary sections). Fractions of which the indices differ a multiple of 15 belong to the same sub-series.

Conjecture 1

For all fractions that belong to the same sub-series, the value 'v' = (numerator)²-n*(denominator)² is the same.

In this example:

Indices	V
0 ± k*15	1	(making these fractions part of the sp-blocks)
1 ± k*15	-13	(making these fractions part of the sp-blocks)
2 ± k*15	-12
3 ± k*15	-9
4 ± k*15	-4
5 ± k*15	3
6 ± k*15	-3
7 ± k*15	4
8 ± k*15	-1
9 ± k*15	12
10 ± k*15	9
11 ± k*15	4
12 ± k*15	-3
13 ± k*15	3
14 ± k*15	-4

If 'v' is negative, the fraction is below the root, otherwise above.
The conjecture holds if indices are extended into the negative realm, hence the ± sign.

The key question
Why are there infinitely many solutions of the diophantine equation s²-n*p²=1?
Or to put it another way, why is the value v = 1, and only the value v=1, an element of the set for every 'n'?
At the core of the set of series that make up the fractional approach of any natural number, is the same 'trivial ab-block':

...	1	0	...
...	0	1	...
-1	0	1	2

For U₀ ⇒ v=1
For U₁ ⇒ v=-n

According to conjecture 1, for every integer 'k' and every natural non-square 'n', the following holds for U_0+kn:
(numerator)²-n(denominator)²=1, so here's the key answer:

There are infinitely many solutions of the diophantine equation s²-n*p²=1
because of the existence of a trivial solution that is independent of 'n'.

The relationship between the two fractions of the sp-block.
For U₁ ⇒ v=-n is a relationship that is also independent of 'n'. Can a relationship be established between U_0+kn and U_1+kn?

Conjecture 2

Of any two sub-series the value (numerator)1*(denominator)2 - (denominator)1*(numerator)2 of two terms in the same section, has the same value in every section.

Example
The above example of √13 may serve as a point in case. Let (U_-26 - U_-17) be an arbitrary choice of terms in the leftmost section, then (U_-11 - U_-2), (U₄ - U₁₃) and (U₁₉ - U₂₈) are the corresponding pairs in the subsequent sections. If you cross multiply and substract as in conjecture 2, you'll find the value for each pair is -43.

For the trivial sp-block, the value in question equals 1. Thus, assuming the truth of conjectures 1 and 2 the following relationships hold for any sp-block with fractions a/b and c/d:

ad-bc = 1

a²-nb² = 1

c²-nd² = -n

From the first one we conclude that c = (ad-1)/b. Substituted in the last we get:

⇒ (a²d²-2ad+1)/b²-nd²+n = 0
⇒ a²d²-2ad+1-nb²d²+nb² = 0
⇒ d²(a²-nb²)-2ad+nb²+1 = 0
⇒ d²-2ad+a² = 0
⇒ (d-a)² = 0 from which:

d = a

c = nb²

Thus conjecture 2 affirms the existence of a set of solutions of the the diophantine equation s²-np² = 1 for any non-square 'n': substitution of the above values in 1) renders 2).

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