Originally Posted by

**wonderboy1953** I *thought* you were claiming that no multigrade of level>2 can be "dotted" with another vector, and create another multigrade of level>2. First I never brought up vectors in any of my posts on this thread. I'm talking about dotting with consecutive numbers 1,2,3,4... (if it's a 2-termed multigrade on both sides of the equal sign, a^n + b^n = c^n + d^n, then I would dot with 1 and 2 with the multigrade arranged from lowest to highest numbers on both sides of the equal sign; if it's a 3-termed multigrade on both sides of the equal sign, a^n + b^n + c^n = d^n + e^n + f^n, then I would dot with 1,2,3 with the multigrade going from lowest to highest numbers on both sides of the equal sign, the number of consecutive numbers I dot with depends on the number of terms on either side of the multigrade).

I should point out that I would add a 0 to a multigrade to make it symmetric because the chances of my observation about equality to the second power improves when you do so. E.g. if you have a multigrade: a^n + b^n = c^n + d^n + e^n, then add a 0 to the left side to make it 0^n + a^n + b^n = c^n + d^n + e^n and dot multiply with 1,2,3 because you've made a three-termed multigrade on both sides of the equal sign.

A demonstration: when you have this bigrade, 4^n + 9^n + 2^n = 8^n + 1^n + 6^n for n = 1,2, rearrange from lowest to highest: 2^n + 4^n + 9^n = 1^n + 6^n + 8^n and dot multiply with 1,2,3 to get:

1 x 2^n + 2 x 4^n + 3 x 9^n = 1 x 1^n + 2 x 6^n + 3 x 8^n, you will see that you get equality for only n = 1.

Another demo: I mentioned earlier about the trigrade 2,9,15,8 and 3,5,14,12. Rearrange the numbers to 2,8,9,15 and 3,5,12,14 and dot multiply with 1,2,3,4 to get: 1 x 2^n + 2 x 8^n + 3 x 9^n + 4 x 15^n =

1 x 3^n + 2 x 5^n + 3 x 12^n + 4 x 14^n which is true for n = 1,2.

I've checked a number of multigrades with dot multiplication on the consecutive numbers and n never goes above 2 to have equality just like Fermat's equation also never has equality when n goes above 2. (btw even if you do have symmetric multigrades with its terms arranged from lowest to highest doesn't guarantee equality for n = 1 or n = 2 which also corresponds to FLT).

Would like to mention to Media_Man that I find 16,4,4,1 interesting and I thank him for his input for using his computer.

I would recommend reading Simon Singh's "Fermat's Enigma" particularly where it was proven that elliptic equations and modular forms are the same

(and I feel that my observation relates to this).