{"blog_url":"https://jeneshicc.hatenadiary.org/","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Fjeneshicc.hatenadiary.org%2Fentry%2F20081206%2F1228548718\" title=\"Problem 131 - \u843d\u66f8\u304d\u3001\u6642\u3005\u843d\u5b66\" class=\"embed-card embed-blogcard\" scrolling=\"no\" frameborder=\"0\" style=\"display: block; width: 100%; height: 190px; max-width: 500px; margin: 10px 0px;\"></iframe>","provider_name":"Hatena Blog","height":"190","blog_title":"\u843d\u66f8\u304d\u3001\u6642\u3005\u843d\u5b66","description":"http://projecteuler.net/index.php?section=problems&id=131 import Number {-- n^3 + n^2 p = n^2 (n+p) = cube p : prime -> gcd n (n+p) = 1 \u3088\u308a n^2 \u3068 n+p \u306f\u5171\u901a\u56e0\u6570\u3092\u6301\u305f\u306a\u3044 \u5f93\u3063\u3066 n = a^3 , n+p = (a+k)^3 \u3068\u304a\u3051\u308b \u3053\u3053\u3067 (a+k)^3 = a^3 + 3 a^2 k + 3 a k^2 + k ^3 = n + k(3a^2+3ak+k^2) \u3088\u3063\u3066\u3001 p = k(3a^2+3ak + k^2)\u3067\u3042\u308b\u3053\u3068\u304c\u5206\u304b\u308b \u305d\u3057\u3066\u3001\u2026","title":"Problem 131","categories":["Project Euler","Haskell"],"url":"https://jeneshicc.hatenadiary.org/entry/20081206/1228548718","author_name":"jeneshicc","author_url":"https://blog.hatena.ne.jp/jeneshicc/","version":"1.0","type":"rich","published":"2008-12-06 16:31:58","provider_url":"https://hatena.blog","width":"100%","image_url":null}