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## Solving s^2-np^2=1

__Why s__

^{2}-np^{2}=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 s

^{2}-n*p

^{2}=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 U_{0} ⇒ v=1For U _{1} ⇒ 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 s
^{2}-n*p^{2}=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 U

_{1}⇒ 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}), (U

_{4}- U

_{13}) and (U

_{19}- U

_{28}) 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
- a
^{2}-nb^{2}= 1 - c
^{2}-nd^{2}= -n

⇒ (a

⇒ a

⇒ d

⇒ d

⇒ (d-a)

^{2}d^{2}-2ad+1)/b^{2}-nd^{2}+n = 0⇒ a

^{2}d^{2}-2ad+1-nb^{2}d^{2}+nb^{2}= 0⇒ d

^{2}(a^{2}-nb^{2})-2ad+nb^{2}+1 = 0⇒ d

^{2}-2ad+a^{2}= 0⇒ (d-a)

^{2}= 0 from which:- d = a
- c = nb
^{2}

^{2}-np

^{2}= 1 for any non-square 'n': substitution of the above values in 1) renders 2).