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#1 Puzzles and Games » Is this another version of Fermat's Last Theorem? » 2009-11-17 06:08:38

wonderboy1953
Replies: 2

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 or k>2 or n>2. The final step is to prove or disprove that this correlates exactly with Fermat's Last Theorem, i.e. one and the same.

#2 Puzzles and Games » How close to Fermat's theorem? » 2009-11-01 09:04:51

wonderboy1953
Replies: 0

MS = Magic Squares(s)

Multigrade example (through to the fifth power, a pentagrade):

1+15+22+50+57+71 = 2+11+27+45+61+70 = 5+6+35+37+66+67 = 216

12+152+222+502+572+712 = 22+112+272+452+612+702

= 52+62+352+372+662+672 = 11,500

13+153+223+503+573+713 = 23+113+273+453+613+703

= 53+63+353+373+663+673 = 682,128

14+154+224+504+574+714 = 24+114+274+454+614+704

= 54+64+354+374+664+674 = 42,502,564

15+155+225+505+575+715 = 25+115+275+455+615+705

= 55+65+355+375+665+675 = 2,724,334,416.

Multigrade equalities still hold by adding any integer n to every base term in both sides. Hence, the base equality always has number 1 appearing in one side (you have to imagine the exponent in this example).

I like recreational math. I was exploring multigrades in connection with MS. I was thinking outside of the box and I thought about doing a dot product with some multigrades.

There's this MS:

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

The first and last rows make a bigrade. I did a dot product with the numbers 1,2,3,4 on the first and last rows and there was equality on the first power, but not the second. The numbers 2,9,15 and 8 make a trigrade with 3,5,14 and 12 so I tried a 1,2,3,4 dot product with this trigrade. There's equality for the first and second powers (I like to refer to this as the first and second levels), but no equality for the third power. Strange. I tried a dot product (1-8) with the trigrade 1,6,11,16,4,7,10,13 and the remaining numbers in the MS, but still no equality beyond the second level (please note that the members of the multigrade must be arranged from low to high).

More recently I tried higher multigrades (tetragrades, pentagrades, etc.), but no equality beyond the second level (btw some multigrades may have no equality when applying the dot product). This situation reminds me of Fermat's theory which was only proven in 1994. If this is true for all cases, then the following questions arise: first, how close is this to Fermat's theorem where you can translate what I call Gauss's dot multiplier (GDM) on the multigrades into Fermat's equation? second, how easy is it to prove that GDM holds for all cases with the multigrades? (Andrew Wiles proof on Fermat's theorem ran 130 pages), third, if true, how much would this add to our understanding of math?

So that's really the story. I don't know what I stumbled onto and maybe someone out there can help out with this.

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