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简友老师的“由勾股定理想到的”(本刊2016第1—2期,以下简称文①)一文中探究了如下问题:正整数n取何值时,代数式(n(n+1))/2的值是完全平方数?简老师通过大量计算,最后归纳得到了一个漂亮的结论:设(n(n+1))/2=k~2(k为正整数),则n的值可由下面的递推公式给出:若n是奇数,n_1=1,n_m=8k_mk_(m+1)+1;若n是偶数,n_m=8k~2_m(m=1,2,3,…).其中k_m=a_(2n-1)a_(2m),a_1=a_2=1,a_(2m+1)=a_(2m-1)+a_(2m),a_(2m+2)=a_(2m-1)+a_(2m+1).笔者发现,该问题的实质是求关于n的一元二次
In his article “The Pythagorean Theorem” (Article 1-2, 2016, hereinafter referred to as ①), Jane Austen explores the following questions: What is the value of the positive integer n, the algebraic expression (n (n + 1)) / 2 is the value of the perfect square. After a lot of calculations, Jane concludes with a pretty good conclusion: Suppose (n (n + 1)) / 2 = k ~ 2 (k is a positive integer) The value of n can be given by the following recurrence formula: If n is odd, n_1 = 1, n_m = 8k_mk_ (m + 1) +1; if n is even, n_m = 8k ~ 2_m (2m) = a_ (2n-1) a_ (2m), a_1 = a_2 = 1, a_ (2m + 1) = a_ (2m-1) + a_ (2m), a_ (2m + 2) = a_ (2m-1) + a_ (2m + 1). The author found that the essence of the problem is to find a quadratic