# Conjecture on triangular numbers T(n)

1. Jan 18, 2008

### ramsey2879

Let a triangular number T(n) = n*(n+1)/2 be factored into the product A*B with A less or equal to B.
Let gcd(x,y) be the greatest common divisor of x and y
For each of pair (A,B) define C,D,E,F as follows

C = (gcd(A,n+1))^2,
D = 2*(gcd(B,n))^2,
E = 2*(gcd(A,n))^2,
F = (gcd(B,n+1))^2.

As an example.
Let n = 36 then T(n) = 666 which can be factored into 6 distinct pairs A,B. The
six sets (A,B,C,D,E,F) are as follows

1. (1,666,1,648, 1369,2)
2. (2,333,1,162,1369,8)
3. (3,222,1,72,1369,18)
4. (6,111,1,18,1369,72)
5. (9,74,1,8,1369,162)
6. (18,37,1,2,1369,648)

My conjecture is that the products (A+Ct)*(B+Dt) and (A+Et)*(B+Ft) are each triangular numbers (= T(r) = r(r+1))/2) for all integer t. Also, the above example gave a interesting result in that when the 12 results for r (= the integer part of the square root of 2T(r)) are sorted by value the difference between adjacent r values when t = 1 was one of the even factors of 666. It would seem to me that the easiest way to prove this conjecture would be to derive and prove a formula for r. I will give the corresponding r values below for t = 1
1. 72,110
2. 54,184
3. 48,258
4. 42,480
5. 40,702
6. 38,1368

Also I found that r follows an arithmetic series as t varies, so it is important to find the pattern for the above values, since the difference between them and 36 is the difference value of the arithmetic series. Anyone see a pattern?

Last edited: Jan 18, 2008
2. Jan 19, 2008

### al-mahed

just a silly question related with you are posting here: is there triangular prime numbers?

I think should be but in a quick search I didn't find an example

3. Jan 20, 2008

### dodo

A formula can be derived from the (positive) solution of r^2 + r - 2*T(r) = 0, which is
$$r = {{\sqrt {1 + 8 T(r)} - 1} \over 2}}$$
But I don't see how this helps to prove your conjecture.

Last edited: Jan 20, 2008
4. Jan 20, 2008

### HallsofIvy

As stated in the original post, any triangular number can be writen in the form n(n+1)/2. Since one of either n or n+1 must be even, that is the product of positive integers. The only way it could be prime is if one of those numbers is 1: if n= 1, then 1(2)/2= 1 which is not a prime number. If (n+1)/2= 1, then n+1= 2 so n= 1 again.

No, there are no prime triangular numbers.

5. Jan 20, 2008

### al-mahed

Yes, this is obvious, I didn't see the most obvious thing !!

6. Jan 20, 2008

### dodo

Actually, T(2)=3 is prime... but the reasoning holds for higher n.

7. Jan 20, 2008

### ramsey2879

Progress report

While only one prime, 3, can be represented as a triangular number, all primes other than 3 can be represented as n(n+1)/2 -i where i is a positive integer less than n in one and only one way, and as another posted noted triangular numbers and square numbers are related since 8T(n) + 1 = (2n+1)^2.

Progress report: my conjecture does indeed hold where n is even and A is a factor of n other than n. First of all I forgot to note that my formula for C,D,E,F was meant only for the case where n is even, if n is odd then make the substitution n' = -(n+1) and the formula also works. I still am working on the general conjecture.

Proof of my simplified conjecture follows:

The new values of A,B,C,D,E, and F can now be expressed as

1A. A,B,1,2*(n/2A)^2,2*A^2,(n+1)^2

The second conjectured result follows directly from the relation

$$T_{(n + 2A(n+1)t)} = (A +2t*(A^2))*(B+ t*(n+1)^2)$$

The first conjectured result (using C and D) follows directly from the relation

$$T_{(n + nt/A)} = (A + t)*(B + 2t*(n/2A)^2)$$

The above relations follow directly by expansion of both sides and comparison of like terms after making the substitution n(n+1)/2 = AB

Although the above is presumably dealing with integers only, I think the same relationship hold for n and A being within the field of rationals or within the complex number field as long as A,B,C,D,E,and F are defined as in line 1A. In this case the conditions that n is odd or that A is a factor of n are not necessary as long as T(n) is defined as n(n+1)/2 regardless of the nature of n.

I will concentrate on A = a factor of n+1 next

Last edited: Jan 21, 2008
8. Jan 22, 2008

### al-mahed

You conjecture is true for A = n/2 and B = n+1, when n = even number, for the expression (A+Dt)(B+Ct)

as gcd(n,n+1) = 1, then C = 1 and D = 2

$$(A+Dt)(B+Ct) = (n/2 + t)(n+1 + 2t) = \frac{[n+2t]*[(n+2t)+1]}{2}$$

But (A+Et)(B+Ft) is not a triangular number for all t

gcd(A,n) = n/2 ==> $$2[gcd(A,n)]^2 = \frac{n^2}{2}$$ ==>

==> $$A+Et = \frac{n + n^2t}{2}$$

gcd(B,n+1) = n+1 ==> $$gcd(B,n+1)^2 = (n+1)^2$$ ==>

==> $$B+Ft = n + 1 + (n+1)^2t = n + n^2t + (2nt + t + 1)$$

showing that (A+Et)(B+Ft) is triangular if and ony if t = 0

Last edited: Jan 22, 2008
9. Jan 22, 2008

### al-mahed

We have to test when A is not n/2, say A = kx and B = wy

10. Jan 22, 2008

### al-mahed

write A = kx and B = wy

its easy to see that if k divides n/2 and x not, then x divides n+1
same argument for B factors

supose we take A and B such that k divides n/2, x does not, w divides n+1, y does not, so n/2 = ky and n+1 = xw ==> 2ky + 1 = xw

gcd(A,n+1) = x ==> C = x^2
gcd(B,n) = y ==> D = 2y^2
gcd(A,n) = k ==> 2k^2
gcd(B,n+1) = w ==> w^2

now we have to find a r(r+1)/2 form

11. Jan 22, 2008

### ramsey2879

You didn't look far enough!
$$(A + Et)*(B+Ft) = \frac{n+n^2t}{2} *(n+n^2t +2nt + t + 1)$$

$$= \frac{n+n^{2}t}{2}*((n+n^{2}t+ nt+1) + nt + t)$$

$$= \frac{n +n^{2}t +nt}{2}*(n+n^{2}t + nt + 1)$$

$$= T_{n + n^{2}t +nt}$$

What's more important you can let A be any integer, rational or complex number and still the product
(A+Et)*(B+Ft) = m(m+1)/2 where T(n) = A*B is a triangular number and E and F are defined as in line 1A of my January 20 post and m = n + 2A(n+1)t

Last edited: Jan 22, 2008
12. Jan 22, 2008

### ramsey2879

Thanks that is helpful, I will use these new terms in my next response.

13. Jan 22, 2008

### ramsey2879

For (A+Ct)*(B + Dt); r = n+2xyt
by symmetry
For (A+Et)*(B+Ft); r = n+2wkt

14. Jan 23, 2008

### ramsey2879

Now put k = (a^n)/2, y = 1/b^n, x = 1/b^n and w = c^n then my equations become

T(2ky + 2xyt) = (kx +tx^2)*(wy+2ty^2)

T(2ky +2kwt) = (ky +2tk^2)*(wy+tw^2)

Is it true that for n>2 this can have no solution except a = 0 since the requirement that 2ky + 1 = wx can not be met due to Fermat's theorm?

Last edited: Jan 23, 2008
15. Jan 23, 2008

### al-mahed

This is not correct, you cannot put "nt" in the other side like this, and what about "t"?

16. Jan 23, 2008

### ramsey2879

(n+n^2t)*(nt+t)=nt*(1+nt)*(n+1) = nt*(n + n^2t + nt + 1) so I was right to express it that way. Sorry for not showing all steps but I had trouble with my latex expressions.

17. Jan 23, 2008

### al-mahed

we have to write (A+Et)*(B+Ft) in the r(r+1)/2 form, right?

if $$A+Et = \frac{n + n^2t}{2}$$, then $$n + n^2t$$ can be called "r"

as$$B+Ft = n + (n+1)^2t + 1$$, and $$n + n^2t \neq n + (n+1)^2t$$, I can NOT write the product (A+Et)*(B+Ft) in the r(r+1)/2 form, since $$n + (n+1)^2t$$ is not "r",

try to find a concrete counter example proving that (A+Et)*(B+Ft) is triangular for t > 0, I may be mistaken, but I believe that there is not such example.

18. Jan 24, 2008

### ramsey2879

I just finished showing that r = n + n^2t + nt . You are under the false impression that r must equal (A+Et) or (B+Ft) but all I said was that r(r+1)/2 = (A+Et)*(B+Ft)
In the example you chose r = n+n^2t +nt . Work it out and see for yourself.

19. Jan 25, 2008

### al-mahed

Sorry, my big mistake, pure laziness !!! Of course you're right

20. Jan 29, 2008

### ramsey2879

Let T(r) = r(r+1)/2
I have used this format as follows
Let n = 2ab and n+1 = cd then the following 4 equations hold for all t
T(n + 2act) = ac(b+ct)(d + 2at)