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).

Originally Posted by wonderboy1953

"I cannot enlighten you with my derivation of this result. I took my own advice and let a computer sift through thousands of candidates and that is one that popped up." You have already enlightened me. I hope I have enlightened you.

BTW I have came up with another method for making a multigrade two days ago starting with another multigrade which I'll demonstrate with the base numbers:

2,8,15,9 = 3,5,14,12 (a trigrade), go from right to left and add numbers that are next to each other - the end numbers add the numbers on the far left:

10^n + 23^n+ 24^n + 11^n = 8^n+19^n + 26^n + 15^n which is valid for n = 1,2,3. Simple algebra will show why this method works.

I want to thank you for helping me out with your computer (incidentally I know about half a dozen methods for making multigrades. I refrain from going into this because I don't know if the purpose of this website is for recreational math or more serious math such as helping out with homework problems - if you read the book I mentioned, you'll see why I posted my findings here as it seems to be of major importance to mathematics)

It can be a year before I get a home computer. In the meantime I look forward to your computer help, but the best thing would be to prove or disprove my conjecture. (thank you too Tonio)