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

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':

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.

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:

1. ad-bc = 1
2. a²-nb² = 1
3. c²-nd² = -n

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

• 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).