Thread: How close to Fermat's theorem?

1. How close to Fermat's theorem?

MS = Magic Squares(s)

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.

2. Diophantine Dizziness

This is very interesting stuff. However, I hate to tell you that it has little or nothing to do with the famed Fermat's Last Theorem. That states that no integer solutions $\displaystyle (x,y,z)$ exist that solve $\displaystyle x^n+y^n=z^n$ when $\displaystyle n>2$. The formulation of this theorem is extremely specific, as by simply adding a fourth term, $\displaystyle x^n+y^n=z^n+w^n$, there are plenty of solutions for many higher values of $\displaystyle n$.

This is much closer to Hilbert's Tenth problem, which asks whether there is a general method of determining if a Diophantine equation has solutions. Unfortunately, the answer is no. That means that by changing or altering a single thing in a previously known Diophantine problem, you create an entirely new problem whose solution has nothing to do with the seemingly similar original problem.

So no, your question probably cannot somehow be tied to Fermat. As for your conjecture, that no multigrade can be dotted by any vector and made into a new multigrade higher than two, you may do better to recruit the help of a computer program to accrue some evidence for such a claim, before attempting to prove it. Counterexamples tend to be easier to find than proofs.

3. Media_ an

"Counterexamples tend to be easier to find than proofs." I hope that someone does find a counterexample, otherwise this conjecture would need a proof. (btw I don't have a home computer and my time is limited at the library)

4. Media_Man - further commentary

I'm a mathematician by inclination, not a scientist meaning I'm not looking for evidence, just proof or disproof. Euler had his prime generating formula which worked for the first 41 numbers. This doesn't mean it would work for the next 41 numbers, nor even the next (which it doesn't).

Currently I don't have access to a private computer which would help speed things up, but even then at rock bottom, no computer can come up with conceiving a proof which only an intelligent creature or being can do, just check cases.

The proposition I put forth so far corresponds to Fermat's theorem. I'll keep trying to disprove my proposition and I welcome others to do so. It's either that or somehow prove it. If proven and if it translates into all Fermat cases, it may simplify the proof for Fermat's theorem and help deepen our understanding of math plus broaden our math knowledge base.

5. Counterexample

Sorry if I sounded dismissive. It's not that this is an easy problem, because it's not. If you were to find an important connection between multigrades and magic squares, that would be a significant discovery.

In this particular case though, I imagine that you based your conjecture on a few dozen hand calculations, when in fact, only one in ten thousand or so might actually end up working. Multigrades are extremely rare so it should come as no surprise that "dotting" one with a random vector should produce a new multigrade. If you dig through the first 20 straws in a hay stack and don't find a needle, that doesn't mean its not there. But if a computer confirms there is no needle in 90% of the haystack, that lends evidence that there may be none to be found. That is what I meant by "evidence."

I do wish you luck on your continued search for connections.

6. Responding to Media_Man

"Multigrades are extremely rare..." Not true. In fact there are several methods for making them including a particular one that's extremely easy to use on any monograde to make a multigrade(s). I still stand by what I posted before.

7. Clarification

As far as their number goes, it turns out that the number of multigrades is infinite (including what I call multiple multigrades involving more than one equal sign).

Again evidence isn't what's needed, proof is.

I haven't read any literature connecting multigrades to magic squares or FLT, but such a connection is what Wonderboy1953 is attempting to find.

11. Responding

First Media-Man:

"If you were to find an important connection between multigrades and magic squares, that would be a significant discovery." The connection is that neither are valid beyond the second power for any positive number (although at this stage, my observation is still a conjecture).

"If you dig through the first 20 straws in a hay stack and don't find a needle, that doesn't mean its not there. But if a computer confirms there is no needle in 90% of the haystack, that lends evidence that there may be none to be found. That is what I meant by "evidence." The problem here is that there are an infinity of multigrades, not just 10, 100, 1,000 or even a million. So I don't know where you and alunw draw the line in the sand (I'd like to ask how you derived the 16,4,4,1 combo that creates another trigrade?).

" I haven't read any literature connecting multigrades to magic squares or FLT." You should read Alfred Posamentier's "Math Charmers" that shows a connection between MS and multigrades (which goes to infinity).

Now alunw:

"...but not putting the ^ to denote taking powers makes your first post hard to understand." sorry about that as I tried to. I'll be more careful in the future.

"...you really need more evidence than this, and some argument to show why your conjecture might be important." Its possible connection to FLT is what makes it important (I would read Fermat's Enigma by Simon Singh which may enlighten you).

"To get mathematicians more interested in what you are talking about it might be a good idea to back up your claim that you have a method for producing multigrades from a monograde." Your wish is my command so I'll offer you an example:

1 + 4 = 2 + 3, add 1 to each member:
2 + 5 = 3 + 4, now switch sides:
3 + 4 = 2 + 5, now combine the first and third equations:
1 + 3 + 4 + 4 = 2 + 2 + 3 + 5, cancel the 3's:
1^n + 4^n + 4^n = 2^n + 2^n + 5^n, valid for n = 1,2

We have just made a bigrade. You can change the numbers to whatever you like and you can repeat the method to make a trigrade and keep repeating the method to however how high you like to go with the multigrades.

Now HallsofIvy:
"Wonderboy1953, it would help if you would define "multigrade". I googled "multigrade" and got a lot of hits about "multigrade oils" but nothing connected with mathematics." Only the oils? What about the classrooms? Try to focus on math multigrades and search through 20 pages with Google.

A multigrade is an equality relationship between two groups of numbers where the sum of their powers is equal for two or more distinct, positive powers.

12. Clarity

I know you are relying on hand calculations in making these observations. I would be more than happy to use a computer to search for a counterexample, if it exists. Obviously, this has no relevance for proving a conjecture, but is the easiest way to disprove it.

Again, what exactly is it that you are conjecturing?

int max=50;
for (int c1=-max; c1<max; c1++){
for (int c2=-max; c2<max; c2++){
for (int c3=-max; c3<max; c3++){
for (int c4=-max; c4<max; c4++){
if ((-c1+4*c2+c3-4*c4==0)
&&(-5*c1*c1+56*c2*c2+29*c3*c3-80*c4*c4==0)
&&(-19*c1*c1*c1+604*c2*c2*c2+631*c3*c3*c3-1216*c4*c4*c4==0)
&&!(c1==c2 && c2==c3 && c3==c4))
System.out.println("c1="+c1+", c2="+c2+", c3="+c3+", c4="+c4);
}}}}

13. Originally Posted by Media_Man

I know you are relying on hand calculations in making these observations. I would be more than happy to use a computer to search for a counterexample, if it exists. Obviously, this has no relevance for proving a conjecture, but is the easiest way to disprove it.

Again, what exactly is it that you are conjecturing?

int max=50;
for (int c1=-max; c1<max; c1++){
for (int c2=-max; c2<max; c2++){
for (int c3=-max; c3<max; c3++){
for (int c4=-max; c4<max; c4++){
if ((-c1+4*c2+c3-4*c4==0)
&&(-5*c1*c1+56*c2*c2+29*c3*c3-80*c4*c4==0)
&&(-19*c1*c1*c1+604*c2*c2*c2+631*c3*c3*c3-1216*c4*c4*c4==0)
&&!(c1==c2 && c2==c3 && c3==c4))
System.out.println("c1="+c1+", c2="+c2+", c3="+c3+", c4="+c4);
}}}}

Well said. I'd add also that anyone really trying to forward in this site some ideas about something he/she thinks may be important related to mathematics will better learn some basic LaTex in this site (just as I did in the last 3 weeks or so) and will make an effort to write down clearly and extensively about his/her ideas, with definitions, examples and etc.

I also can't tell so far what exactly is wonderboy trying to tell us, let alone what his conjectures about anything are.

Tonio

14. Media_Man (and Tonio)

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).

15. To Media_Man

"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)

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