1. ## from multigrades to GDM to FLT

A multigrade is an equality relationship between two (or more) groups of numbers where the sum of their powers is equal for two or more distinct, positive powers (I first became aware of them through Albert Beiler's "Recreations In The Theory Of Numbers").

Multigrades can be asymmetric or symmetric with respect to their terms. An example of an asymmetric multigrade is 3^n + 3^n = 1^n + 1^n + 4^n which is true for n = 1,2. From all the examples of I've examined, asymmetric multigrades differ by at most one term so for the present example, you have two terms on one side of the equation and three terms on the other side.

However asymmetry is only is a mirage because all asymmetric multigrades can be converted into symmetric ones through the addition of 0. So by adding 0, the above equation is changed into 0^n + 3^n + 3^n = 1^n +1^n + 4^n, n = 1,2.

GDM means Gauss's Dot Multiplier meaning you're dot multiplying by consecutive numbers, starting from 1 on the left side of the equation and dot multiplying again, starting from 1, on the right side of the equation. So for the top equation, multiplying by GDM yields 1 x 3^n + 2 x 3^n for the left side of the equation and 1 x 1^n + 2 x 1^n + 3 x 4^n on the right side of the equation. Note that there is no equality for any value of n.

The importance of the 0 term will now be shown. Plug in 0 to get the second equation above and apply GDM to the second equation to get:

1 x 0^n + 2 x 3^n + 3 x 3^n = 1 x 1^n + 2 x 1^n + 3 x 4^n which is true for n = 1. With trigrades, I've found n = 1,2 can work when I applied the GDM (but there is no guarantee that applying GDM will work with all multigrades).

In July I started to go to higher multigrades (tetragrades, pentagrades, etc.), but n never exceeded 2 after I applied the GDM. This made me think of FLT which I believe Media_Man and Tonio is beginning to understand.

There is much more to multigrades. You can have multigrades with more than one equal sign. You can have alternating multigrades which are true for either even or odd powers of n. I have an example of a multigrade which is true for n = 2,3,4, but not 1. I suspect you can have multigrades that skip powers in other ways (e.g. n = 1,4,7,10...).

I hope in general that all visitors can see what I'm getting at. I'll be dropping from by to by. Thank you you Media_Man for your computer help
as two hours a day at the public library doesn't cut it and thank you Tonio for looking further. I'll do one more thing on this thread.

Take the MS:

16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1

The magic sum is 34. Take the diagonals 3,9,11 and 6,8,14. When you double the middle numbers, you get 3,11,11,9 and 6,6,8,14 which is a bigrade. Have fun

2. ## Very odd indeed

Starting from 2^n + 8^n + 9^n + 15^n = 3^n + 5^n + 12^n + 14^n for n = 1,2,3; apply GDM 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 for n = 1,2; multiply both sides of this equation by 2 to get:

2(1 x 2^n + 2 x 8^n + 3 x 9^n + 4 x 15^n) = 2(1 x 3^n + 2 x 5^n + 3 x 12^n + 4 x 14^n) for n = 1,2; break open parentheses and multiply through to get:

2 x 2^n + 4 x 8^n + 6 x 9^n + 8 x 15^n = 2 x 3^n + 4 x 5^n + 6 x 12^n + 8 x 12^n for n = 1,2; now subtract the first equation from the last to get:

1 x 2^n + 3 x 8^n + 5 x 9^n + 7 x 15^n = 1 x 3^n + 3 x 5^n + 5 x 12^n + 7 x 14^n for n = 1,2

So if GDM works with a multigrade (which is dot multiplication by 1,2,3,4 in this case) then you can convert the coefficients by simple algebraic manipulation into odd numbers, or 1,4,7,10 or 4,3,2,1 or any consecutive series to dot multiply with as long as the multigrade is symmetric being dot multiplied by the same number of terms as in the multigrade which may produce a result that can work for n = 1 or n = 1,2 (but as my conjecture or proposition asserts, never works for n>2).

3. ## Making this more understandable

First, to clarify, I think I posted here in number theory section that I first made my observation back in July so let me correct this. This observation came to me in September after I started looking past trigrades with higher multigrades such as tetragrades, pentagrades, etc.

Second, with multigrades, there's always one term that dominates the others meaning that this term's base is greater than the other terms' base so when the exponent n is high enough, then this term will dominate the other side of the equation. E.g. say a^n + b^n + c^n = d^n + e^n + f^n and f>a, f>b, f>c, then at some point f^n will be > a^n + b^n + c^n for some n and will stay that way as n increases ad infinitum. So, in general, there are an infinite amount of multigrades, but for a particular multigrade, it must stop at some exponent n.

To explain my observation, it's already known that FLT is true. A way of explaining this is to compare with a staircase. Most staircases are 13 steps while some have more steps and some have less. FLT is saying whatever the size of the staircase, you can climb one stair or two stairs or maybe none at all depending on the situation, but you can never climb more than two stairs on the staircase, no matter which staircase it is (and there are an infinite number of staircases in the math universe) and I'm asserting the same holds true with my conjecture.

If my conjecture is proven, the next step is to see if it correlates with FLT for all possible cases. Time will tell.

4. ## Further info

Andrew Wiles proved FLT by first applying Galois's group theory which transformed FLT from being an infinite problem to a finite problem (as described by Amir Aczel's book, "Fermat's Last Theorem."

It may be fruitful to apply Galois's group theory to my conjecture as well to prove or disprove the conjecture.

5^n + 8^n + 9^n + 12^n = 6^n + 6^n + 11^n + 11^n, n = 1,2,3 with repeating base number 6 and 11.

I've never seen a multigrade with repeating base numbers where a repeating base number was the highest term (which for this trigrade is 12). This doesn't mean that such a multigrade doesn't exist; if such a multigrade does exist and you apply GDM to it, would you get a combo GDM-multigrade with equality on both sides of the equal sign for some exponent and can the exponent exceed 2 and still have equality?

I ordered a book written by Amir Aczel titled Fermat's Last Theorem. After I read it at my leisure, I'll return to this website to see what's up (probably next week).

6. ## New info on a previous post

On some thread I posted from the MS:

16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1

that you can derive the trigrade:

2^n + 8^n + 9^n + 15^n = 3^n + 5^n + 12^n + 14^n for n = 1,2,3 (the base numbers from either side of the equal sign form lopsided squares within the MS). Next I applied GDM to derive:

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 for n = 1,2 meaning while n works for 1 - 3 for the trigrade, it only works for 1 and 2 for GDM applied to this trigrade.

It recently occurred to me that if GDM works on the multigrade, what would happen if I reverse the situation by dot multiply the base numbers by GDM raised to a power. So I successfully tried this today and I got:

1^n x 2 + 2^n x 8 + 3^n x 9 + 4^n x 15 = 1^n x 3 + 2^n x 5 + 3^n x 12 + 4^n x 14 which works for n = 1,2.

This is the only case I've tried this out for, however I'm willing to bet that where applying GDM to any multigrade works for n = 1 or n = 1,2, then by applying GDM raised to a power on the base numbers within the multigrade will also work for n = 1 or n =1,2 (in fact there's another multigrade within that MS where you can form a trigrade using the diagonals 1,4,6,7,10,11,13,16 with 2,3,5,8,9,12,14,15 and dot multiply with 1 - 8 and there are other multigrades as well).

I'm not aware of any other studies done in this area let alone a name for the process of this special dot multiplication. So I'm going to call the result of this special dot multiplication process a hypermultigrade (including dot multiplying the base numbers of the multigrade by GDM raised by a power).

7. ## Extending my conjecture

I've been studying the hypermultigrade 1^k x a^n + 2^k x b^n + 3^k x c^n... = 1^k x d^n + 2^k x e^n + 3^k x f^n.... It appears that this creature (the hypermultigrade) will never work when (k + n)>3 and can work when 0<=(k + n)=<3 [with the sole exception that a base term nor a coefficient can't equal 0 when (k + n) = 0] which is my extended conjecture. The next step is to determine which types of multigrades it will work for.

The intermediate step is to prove that the equation on top won't be valid in whole numbers whenever (k + n)>3. The final step is to prove or disprove that this correlates exactly with Fermat's Last Theorem, i.e. one and the same.

8. MHF is not the place to be posting your theories and work in progress. Also, it's not appropriate to, effectively, be spamming this thread with defective latex code while you practice latex.