λ＞3のペル数を扱いたいのであるが、ax^2-bx-c=0

λ=√(9・a^2+4)/a

解の大きいほうが3より大きくなければならない。

{-ｂ+√(9・a^2+4)}/2a＞3

-ｂ＞6a-√(9・a^2+4)

|b|は3aよりおおくならなければならない。

a=2のとき2x^2-bx+c

|b|=6,6^2-4・2・c＝40

|b|=7,7^2-4・2・c＝40

|b|=8,8^2-4・2・c＝40,C=3

a=12のとき12x^2-bx+c

|b|=36,36^2-4・12・c＝1300

|b|=37,37^2-4・12・c＝1300

|b|=38,38^2-4・12・c＝1300,c=3

a=70のとき70x^2-bx+c

|b|=210,210^2-4・70・c＝44104

|b|=211,211^2-4・70・c＝44104

|b|=212,212^2-4・70・c＝1300,c=3

(3a+2)^2-4a・3=(9a^2+4)が成り立つ

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√5/1→x^2-x-1

√8/2→x^2-2x-1

√221/5→5x^2-11x-5

√1517/13→13x^2-29x-13

√7565/29→29x^2-63x-31・・・ペル

√10400/34→17x^2-38x-17

√71285/89→89x^2-199x-89

√257045/169→169x^2-367x-181・・・ペル

√338720/194→97x^2-216x-98・・・どちらでもない

√488597/233→233x^2-521x-233

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λ>3に対して

√13/1→x^2-3x-1=0

√85/3→3x^2-7x-3=0

√580/8→8x^2-18x-8=0=0

√3973/21→21x^2-47x-21=0

√27229/55→55x^2-123x-55=0

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2x^2-8x+3=0

x=(4+√ 10)/2

x=3+{-2+√10)}/2+++

x=3+1/(2/{-2+√ (10)})

x=3+1/({2+√ 10)}/3)

x=3+1/(1+{-1+√ (10)}/3)

x=3+1/(1+1/(3/{-1+√ (10)})

x=3+1/(1+1/({1+√ 10)}/3)

x=3+1/(1+1/(1+{-2+√ (10)}/3)

x=3+1/(1+1/(1+1/(3/{-2+√ (10)})

x=3+1/(1+1/(1+1/({2+√ (10)}/2)

x=3+1/(1+1/(1+1/(2+{-2+√ 10)}/2)+++

x=3+1/(1+1/(1+1/(2+1/(2/{-2+√ (10)})

x=3+1/(1+1/(1+1/(2+1/({2+√ (10)}/3)

x=3+1/(1+1/(1+1/(2+1/(1+{-1+√ (10)}/3)

x=3+1/(1+1/(1+1/(2+1/(1+1/(3/{-1+√(10)})

x=3+1/(1+1/(1+1/(2+1/(1+1/({1+√(10)}/3)

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+{-2+√ (10)}/3)

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(3/{-2+√(10)})

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/({2+√10)}/2))

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(2+{-2+√(10)}/2))+++

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(2+1/(2/{-2+√(10)})

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(2+1/{2+√ (10)}/3)

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(2+1/(1+{-1+√ 10)}/3)

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(2+1/(1+1/(3/{-1+√(10)})

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(2+1/(1+1/{1+√ (10)})/3}

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(2+1/(1+1/(1+{-2+√ (10)})/3}

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(2+1/(1+1/(1+1/(3/{-2+√ 10)})

x=3+1/(1+1/(1+1/(2+1/(1+1/(1+1/(2+1/(1+1/(1+1/({2+√(10)}/2)

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x=[3:1,1,2,---1,1,2---,2,1,1,1,1,2,2,1,1,2,・・・]

x-3=1/(1+(1/1+1/(2+(x-3))

x-3=1/(1+(1/1+1/(x-1))

x-3=1/(1+(x-1)/(x))

(x-3)=x/(2x-1)

(x-3)(2x-1)=x

2x^2-7x+3=x・・・（OK)

しかし

y=(-4+√ 10)/2・・・負になってしまう。ペル型は存在しないのかもしれない。

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a=2のとき2x^2-8x+3

x=(8+√40)/4＞3

y=(-8+√40)/4＜0

a=12のとき12x^2-38x+3

x=(38+√1300)/24＞3

y=(-38+√1300)/24＜

a=70のとき70x^2-212x+3

x=(212+√44104)/140＞3

y=(-212+√44104)/140＜0

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