### Who's Online

We have 129 guests and no members online

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 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.

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: