# Arithmetic Series and Triangular Numbers

by ramsey2879
Tags: arithmetic, numbers, series, triangular
 P: 891 Re: Arithmetic Series and Factors of Triangular Numbers A+C*n, B+F*n, A+E*n, B+D*n are all arithmetic series which I define below It is still not clear to me as why (A+C*n)*(B+F*n) and (A+E*n)*(B+D*n) are both triangular numbers for all integers n Can someone please visit my blog and explain it? Perhaps someone can show somehow that $$8*(A+C*n)*(B+F*n) + 1$$ and $$8*(A+E*n)*(B+D*n) + 1$$ are both perfect squares for all n using Mathematica or some other program. --- In Triangular_and_Fibonacci_Numbers@yahoogroups.com, "ramsey2879" wrote: > > I found a previously unknow property of Triangular Numbers AFAIK. > Given that T is a triangular number having exactly M distinct ways to > pair the product into two factors, A*B with A M distinct pairs, there are two coprime pairs of integers (C,E) and > (E,D)where C*D = the perfect square of integral part of the square root > of 2*T and E*F = the next higher perfect square and each of the > products (A+Cn)*(B+Fn) and (A+En)*(B+Dn) are triangular numbers for all > integer values of n. > As an example. > Let T = 666 which can be factored into 6 distinct pairs A,B. The six > sets (A,B,C,D,E,F) are as follows > > 1. (1,666,1,1369,2,648) > 2. (2,333,1,1369,8,162) > 3. (3,222,1,1369,18,72) > 4. (6,111,1,1369,72,18) > 5. (9,74,1,1369,162,8) > 6. (18,37,1,1369,648,2) > > I don't have a proof of the general result but am working on it. > Still no proof, however, let ab = T(r) = r(r+1)/2, gcd(n,m) = the greatest common divisor of m and n, then the formula for C,D,E and F as a function of A and B is Case 1, r is even C = (gcd(A,r+1))^2, F = 2*(gcd(B,r))^2 E = 2*(gcd(A,r))^2, D = (gcd(B,r+1))^2 The determinant |C F| |E D| = (r+1)^2 - r^2 = 2r+1 Case 2 r is odd C = 2*(gcd(A,r+1))^2, F = (gcd(B,r))^2 E = (gcd(A,r))^2, D = 2*(gcd(B,r+1))^2 The determinant |C F| |E D| = (r+1)^2 - r^2 = 2r+1

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