Let T(n)=n(n+1)/2. As discovered in my earlier efforts an expanded(adsbygoogle = window.adsbygoogle || []).push({});

representation of the values for A-F that solve the formula T(A*n+B)=

(C*n+D)*(E*n+F), for all integer n is as follows:

A= 4t^2*m^2 + 6t^2*m + 2t^2

B= (2t^2-2t)*m^2 + (3t^2-3t-2)*m +(t^2-t-2)

C= 2t^2*m^2 + 4t^2*m + 2t^2

D= (t^2-t)*m^2 + (2t^2-2t-1)*m + (t^2-t-1)

E= 4t^2*m^2 + 4t^2*m + t^2

F= (2t^2-t)*m^2 + (2t^2-2t-2)*m + (t^2-t-2)/2

I noted that n can be any real number(I suspect that n could even be

complex!!), for instance if m=2,t=2 and n=1/3 the equation T(A*n+B)=

(C*n+D)*(E*n+F) becomes

T({8/3}*{8+6+1}+16+8) = ({8/3}*{4+4+1}+8+6+1)*({4/3}*{16+8+1}+16+4)

=T(64)=39*(4/3*25 + 20) = 13*(100+60) = 65*32 = 64*(64+1)/2

Notice how the factors 13 and 5 of 65 conveniently become separated

between the two factors of T(64). This method could also work if we let

A*n +B be one less than any other number that needs factoring.

As an example of factoring 63 which is a Mersenne number, we can try

letting A*n+B = -64 by setting n = -11/15. Then my tri-equation becomes

T(-64)=-189/5 * -160/3 = 63*64/2 but this didn't factor the 63.

However, let T=2,M=3,n=1/2 then my tri-equation becomes

T(-64)= 36*56 which sucessfully factors the 63 as one merely has to divide each factor by the factors 4,8 of 64/2.

I believe this method has a lot of potential. Any comments

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Factoring Mersenne and other numbers using tri-math

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**