\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n
(1)<\/strong>\u3000\\(n\\) \u3092\u6b63\u306e\u6574\u6570, \\(a = 2^n\\) \u3068\u3059\u308b. \\(3^a-1\\) \u306f \\(2^{n+2}\\) \u3067\u5272\u308a\u5207\u308c\u308b\u304c \\(2^{n+3}\\) \u3067\u306f\u5272\u308a\u5207\u308c\u306a\u3044\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(m\\) \u3092\u6b63\u306e\u5076\u6570\u3068\u3059\u308b. \\(3^m-1\\) \u304c \\(2^m\\) \u3067\u5272\u308a\u5207\u308c\u308b\u306a\u3089\u3070 \\(m = 2\\) \u307e\u305f\u306f \\(m = 4\\) \u3067\u3042\u308b\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n 1*<\/strong>\u3000\\(n=1\\) \u306e\u3068\u304d\r\n\\[\r\n3^a -1 = 3^2 -1 = 8 = 2^3\r\n\\]\r\n\u306a\u306e\u3067 \\(n=1\\) \u306e\u3068\u304d, [\uff0a] \u306f\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n 2*<\/strong>\u3000\\(n=k \\ ( k \\geqq 1 )\\) \u306e\u3068\u304d [\uff0a] \u304c\u6210\u7acb\u3059\u308b, \u3059\u306a\u308f\u3061\r\n\\[\r\n3^{2^k}-1 = 2^{k+2} M \\quad ( M \\text{\u306f\u5947\u6570} ) \\quad ... [1]\r\n\\]\r\n\u3068\u4eee\u5b9a\u3059\u308b. 1*<\/strong> 2*<\/strong>\u3088\u308a, \u984c\u610f\u306f\u793a\u3055\u308c\u305f.<\/p>\r\n (2)<\/strong><\/p>\r\n \\(m = 2^{\\ell} L\\) \uff08 \\(L\\) \u306f\u5947\u6570\uff09\u3068\u304a\u304f.\r\n\\[\\begin{align}\r\n3^m -1 & = 3^{2^{\\ell} L} -1 = \\left( 3^{2^{\\ell}} \\right)^L -1 \\\\\r\n& = \\underline{ \\left( 3^{2^{\\ell}} -1 \\right) } _ {[A]} \\underline{ \\left\\{ \\left( 3^{2^{\\ell}} \\right)^{L-1} + \\cdots +3^{2^{\\ell}} +1 \\right\\} } _ {[B]}\r\n\\end{align}\\]\r\n (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a, \u4e0b\u7dda\u90e8 [A] \u306f \\(2^{\\ell +2}\\) \u3067\u5272\u308a\u5207\u308c\u308b\u304c, \\(2^{\\ell +3}\\) \u3067\u306f\u5272\u308a\u5207\u308c\u306a\u3044. 1*<\/strong>\u3000\\(\\ell = 1\\) \u306e\u3068\u304d, [4] \u3088\u308a\r\n\\[\r\n3 \\geqq 2^1 L\r\n\\]\r\n\\(L\\) \u306f\u5947\u6570\u306a\u306e\u3067\r\n\\[\r\nL=1\r\n\\]\r\n\u3053\u306e\u3068\u304d\r\n\\[\r\nm = 2^1 \\cdot 1 = 2\r\n\\]<\/li>\r\n 2*<\/strong>\u3000\\(\\ell = 2\\) \u306e\u3068\u304d, [4] \u3088\u308a\r\n\\[\r\n4 \\geqq 2^2 L\r\n\\]\r\n\\(L\\) \u306f\u5947\u6570\u306a\u306e\u3067\r\n\\[\r\nL=1\r\n\\]\r\n\u3053\u306e\u3068\u304d\r\n\\[\r\nm = 2^2 \\cdot 1 = 4\r\n\\]<\/li>\r\n<\/ol>\r\n 1*<\/strong> 2*<\/strong>\u3088\u308a, \u6761\u4ef6\u3092\u6e80\u305f\u3059 \\(m\\) \u306f\r\n\\[\r\nm = 2 , 4\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088. (1)\u3000\\(n\\) \u3092\u6b63\u306e\u6574\u6570, \\(a = 2^n\\) \u3068\u3059\u308b. \\(3^a-1\\) \u306f \\(2^{n+2}\\) \u3067\u5272\u308a\u5207\u308c\u308b\u304c \\(2^{n+3}\\) \u3067\u306f\u5272\u308a\u5207\u308c\u306a\u3044\u3053\u3068\u3092\u793a\u305b. (2)\u3000\\(m … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
\r\n
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\r\n\u3053\u306e\u3068\u304d\r\n\\[\\begin{align}\r\n3^{2^{k+1}}-1 & = 3^{2^k \\cdot 2} -1 \\\\\r\n& = \\left( 3^{2^k} -1 \\right) \\left( 3^{2^k} +1 \\right) \\\\\r\n& = 2^{k+2} M \\cdot \\left( 2^{k+2} M +2 \\right) \\quad ( \\ \\text{\u2235} \\ [1] \\ ) \\\\\r\n& = 2^{k+2} M \\cdot 2 \\left( 2^{k+1} M +1 \\right) \\\\\r\n& = 2^{k+3} M \\left( 2^{k+1} M +1 \\right) \\quad ... [2]\r\n\\end{align}\\]\r\n\\(M\\) , \\(2^{k+1} M +1\\) \u306f\u3068\u3082\u306b\u5947\u6570\u306a\u306e\u3067, [2] \u306f \\(2^{k+3}\\) \u3067\u5272\u308a\u5207\u308c\u308b\u304c, \\(2^{k+4}\\) \u3067\u306f\u5272\u308a\u5207\u308c\u306a\u3044.
\r\n\u3057\u305f\u304c\u3063\u3066, \\(n =k+1\\) \u306e\u3068\u304d\u3082 [\uff0a] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n
\r\n\u307e\u305f, \u4e0b\u7dda\u90e8 [B] \u306f \\(3\\) \u306e\u7d2f\u4e57\u306e \\(L\\) \uff08\u5947\u6570\uff09\u500b\u306e\u548c\u306a\u306e\u3067, \u5947\u6570\u3067\u3042\u308b.
\r\n\u3057\u305f\u304c\u3063\u3066, \\(3^m -1 = 2^{\\ell +2} I\\) \uff08 \\(I\\) \u306f\u5947\u6570\uff09\u3068\u8868\u305b\u308b.
\r\n\u3053\u308c\u304c \\(2^{m} = 2^{2^{\\ell} L}\\) \u3067\u5272\u308a\u5207\u308c\u308b\u6761\u4ef6\u306f,\r\n\\[\r\n\\ell +2 \\geqq 2^{\\ell} L \\quad... [4]\r\n\\]\r\n\\(\\ell \\geqq 3\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\n2^{\\ell} & = ( 1+1 )^{\\ell} = \\textstyle\\sum\\limits _ {i=0}^{\\ell} {} _ {\\ell} \\text{C} {} _ i \\cdot 1^i \\cdot 1^{\\ell -i} \\\\\r\n& = 1 +\\ell + \\cdots +\\ell +1 \\gt \\ell +2\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \\(\\ell = 1 , 2\\) \u306e\u3068\u304d\u306b\u3064\u3044\u3066\u8003\u3048\u308c\u3070\u3088\u3044.<\/p>\r\n\r\n