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.
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 U0+kn:
(numerator)2-n(denominator)2=1, so here's the key answer:
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?
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:
⇒ (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: