# Arithmetic Series and Triangular Numbers

1. Sep 18, 2007

### ramsey2879

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"
<ramseykk2@...> 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 </= B. For each of
these
> 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