Editing Pell number (section)

=== ''H''<sup>2</sup>&nbsp;−&nbsp;2''P''<sup>2</sup>&nbsp;=&nbsp;±1 ===

Since {{sqrt|2}} is irrational, we cannot have ''{{sfrac|H|P}}''&nbsp;=&nbsp;{{sqrt|2}}, i.e.,

:<math>\frac{H^2}{P^2} = \frac{2P^2}{P^2}.</math>

The best we can achieve is either

:<math>\frac{H^2}{P^2} = \frac{2P^2-1}{P^2}\quad \mbox{or} \quad \frac{H^2}{P^2} = \frac{2P^2+1}{P^2}.</math>

The (non-negative) solutions to {{nowrap|1=''H''<sup>2</sup> − 2''P''<sup>2</sup> = 1}} are exactly the pairs {{nowrap|(''H<sub>n</sub>'', ''P<sub>n</sub>'')}} with ''n'' even, and the solutions to {{nowrap|1=''H''<sup>2</sup> − 2''P''<sup>2</sup> = −1}} are exactly the pairs {{nowrap|(''H<sub>n</sub>'', ''P<sub>n</sub>'')}} with ''n'' odd. To see this, note first that

:<math>H_{n+1}^2-2P_{n+1}^2 = \left(H_n+2P_n\right)^2-2\left(H_n+P_n\right)^2 = -\left(H_n^2-2P_n^2\right),</math>

so that these differences, starting with {{nowrap|1=''H''{{su|b=0|p=2}} − 2''P''{{su|b=0|p=2}} = 1}}, are alternately 
1 and −1. Then note that every positive solution comes in this way from a solution with smaller integers since 

:<math>(2P-H)^2-2(H-P)^2 = -\left(H^2-2P^2\right).</math>

The smaller solution also has positive integers, with the one exception: {{nowrap|1=''H'' = ''P'' = 1}} which comes from ''H''<sub>0</sub>&nbsp;=&nbsp;1 and ''P''<sub>0</sub>&nbsp;=&nbsp;0.