\u81ea\u7136\u6570\u306e\u6570\u5217 \\(\\{ a _ n \\} , \\{ b _ n \\}\\) \u306f\r\n\\[\r\n\\left( 5+\\sqrt{2} \\right)^n = a _ n +b _ n \\sqrt{2} \\quad ( n = 1, 2, 3, \\cdots )\r\n\\]\r\n\u3092\u6e80\u305f\u3059\u3082\u306e\u3068\u3059\u308b.<\/p>\r\n
(1)<\/strong>\u3000\\(\\sqrt{2}\\) \u306f\u7121\u7406\u6570\u3067\u3042\u308b\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(a _ {n+1} , b _ {n+1}\\) \u3092 \\(a _ n , b _ n\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n (3)<\/strong>\u3000\u3059\u3079\u3066\u306e\u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066, \\(a _ {n+1} +pb _ {n+1} =q \\left( a _ n +pb _ n \\right)\\) \u304c\u6210\u308a\u7acb\u3064\u3088\u3046\u306a\u5b9a\u6570 \\(p , q\\) \u3092 \\(2\\) \u7d44\u6c42\u3081\u3088.<\/p><\/li>\r\n (4)<\/strong>\u3000\\(a _ n , b _ n\\) \u3092 \\(n\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\(\\sqrt{2} = \\dfrac{m}{n}\\) \uff08 \\(m\\) , \\(n\\) \u306f\u3068\u3082\u306b\u81ea\u7136\u6570\u3067\u4e92\u3044\u306b\u7d20 ... [1] \uff09\u3068\u4eee\u5b9a\u3059\u308b\u3068\r\n\\[\r\n2 n^2 =m^2\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(m\\) \u306f \\(2\\) \u306e\u500d\u6570\u3067\u3042\u308a, \\(m = 2m'\\) \u3068\u304a\u304f\u3068\r\n\\[\\begin{align}\r\n2n^2 & = \\left( 2m' \\right)^2 \\\\\r\n\\text{\u2234} \\quad n^2 & = 2{m'}^2\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \\(n\\) \u3082 \\(2\\) \u306e\u500d\u6570\u3068\u306a\u308b\u304c, \u3053\u308c\u306f [1] \u306b\u77db\u76fe\u3059\u308b. (2)<\/strong><\/p>\r\n \u6761\u4ef6\u3088\u308a\r\n\\[\\begin{align}\r\na _ {n+1} +b _ {n+1} \\sqrt{2} & = \\left( a _ n +b _ n \\sqrt{2} \\right) \\left( 5+\\sqrt{2} \\right) \\\\\r\n& = \\left( 5 a _ n +2 b _ n \\right) +\\left( a _ n + 5 b _ n \\right) \\sqrt{2}\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\na _ {n+1} = \\underline{5 a _ n +2 b _ n} , \\ b _ {n+1} = \\underline{a _ n + 5 b _ n}\r\n\\]\r\n (3)<\/strong><\/p>\r\n \\[\\begin{align}\r\na _ {n+1} +p b _ {n+1} & = 5 a _ n +2 b _ n +p \\left( a _ n + 5 b _ n \\right) \\\\\r\n& = (p+5) a _ {n} +(5p+2) b _ n\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\r\n\\left\\{ \\begin{array}{l} q =p+5 \\\\ pq =5p+2 \\end{array} \\right.\r\n\\]\r\n\u3086\u3048\u306b\r\n\\[\\begin{align}\r\np (p+5) & = 5p+2 \\\\\r\np^2 & = 2 \\\\\r\n\\text{\u2234} \\quad p & = \\pm \\sqrt{2} \\\\\r\n\\text{\u2234} \\quad q & = 5 \\pm \\sqrt{2}\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\n( p , q ) = \\underline{\\left( \\pm \\sqrt{2} , 5 \\pm \\sqrt{2} \\right) \\quad ( \\text{\u8907\u53f7\u540c\u9806})}\r\n\\]\r\n (4)<\/strong><\/p>\r\n (3)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\\begin{align}\r\na _ {n+1} +\\sqrt{2} b _ {n+1} & = \\left( 5+\\sqrt{2} \\right) \\left( a _ n +\\sqrt{2} b _ n \\right) , \\\\\r\na _ {n+1} -\\sqrt{2} b _ {n+1} & = \\left( 5-\\sqrt{2} \\right) \\left( a _ n -\\sqrt{2} b _ n \\right)\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(a _ 1 +\\sqrt{2} b _ 1 = 5 +\\sqrt{2}\\) , \\(a _ 1 -\\sqrt{2} b _ 1 = 5 -\\sqrt{2}\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\na _ n +\\sqrt{2} b _ n & = \\left( 5+\\sqrt{2} \\right) \\left( 5+\\sqrt{2} \\right)^{n-1} =\\left( 5+\\sqrt{2} \\right)^n \\quad ... [2] , \\\\\r\na _ n -\\sqrt{2} b _ n & = \\left( 5-\\sqrt{2} \\right) \\left( 5-\\sqrt{2} \\right)^{n-1} =\\left( 5-\\sqrt{2} \\right)^n \\quad ... [3]\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, \\(( \\text{[2]}+\\text{[3]} ) \\div 2\\) , \\(( \\text{[2]}-\\text{[3]} ) \\div 2 \\sqrt{2}\\) \u3088\u308a\r\n\\[\\begin{align}\r\na _ n & = \\underline{\\dfrac{1}{2} \\left\\{ \\left( 5+\\sqrt{2} \\right)^n +\\left( 5-\\sqrt{2} \\right)^n \\right\\}} , \\\\\r\nb _ n & = \\underline{\\dfrac{\\sqrt{2}}{4} \\left\\{ \\left( 5+\\sqrt{2} \\right)^n -\\left( 5-\\sqrt{2} \\right)^n \\right\\}}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u81ea\u7136\u6570\u306e\u6570\u5217 \\(\\{ a _ n \\} , \\{ b _ n \\}\\) \u306f \\[ \\left( 5+\\sqrt{2} \\right)^n = a _ n +b _ n \\sqrt{2} \\quad ( n = 1, 2 … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
\r\n\u3088\u3063\u3066, \\(\\sqrt{2}\\) \u306f\u6709\u7406\u6570\u3067\u306f\u306a\u304f, \u7121\u7406\u6570\u3067\u3042\u308b.<\/p>\r\n