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:
0 ± k*151(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*153
6 ± k*15-3
7 ± k*154
8 ± k*15-1
9 ± k*1512
10 ± k*159
11 ± k*154
12 ± k*15-3
13 ± k*153
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':
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 U0+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 U0+kn and U1+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.

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:
  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:
⇒ (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).