Fermat's last theorum

[Peter van Velzen]

I made a funny spreadsheet today. I checked the equation x cubed + y cubed = z cubed. (the smallest possibility Fermat’s theorem says, cannot work) But I only checked de last digit! I actually left out he possibility 0 0 0, as that doesn’t satisfy my wish to check only triples that had no common factor.

I did it first in the binary mode, and lo and behold only 375 of a thousand possibilities had correct last digits in base 2. In base 3 I got a bit less: namely 271 and 101 possibilities had a correct last digit in both bases. Base four yielded only 190 triplet, of which 52 also appeared in the previous sheets. The first great washout came in base 5. Although there were still 192 triplets in that base, only 19 of them appeared every previous time. A great disappointment (but I should have known that) was base 6 although the count was now down to 100, all 13 previous set were in it and I made no progress. The prime 7 worked a lot better as was to be expected. Though 191 possibilities appeared only 2 of them were in all of the previous sheets. Could base 8 do the trick? Yes it could! The 100 hits I got there did not include the 2 diehearts from the other bases.

I guess I proved the cubic variant is not solvable. If you can’t get the last digit right you do not have a decent answer! This went better than expected; I was counting on losing about half of the leftovers from each base, and sort of expected to have to go to base 10. Or – if I was unlucky - even beyond.. In which case I gues I would have to expand my testset from 1000 to 1331 or more. I could sort of predict base 2 and 3 as I had already calculated they would give 3 good triplets out of 8 and 8 good triplets out of 27 respectively, but beyond that the third power crept in, by changing the last digit to a smaller set of possibilities. (either less or more plausible) The 2’s all disappeared (…2 x …2 = …0 in base four, and once your last digit is zero, it never ever changes to anything else no matter what you multiply it with. So from there on I hadn’t predicted the results I got.
Maybe I will give the map a try and the fifth power next. Power 4 I can actually prove is impossible, so that’s no fun.

[pallieter]

Salander began her advance towards the house, moving in a circle through the woods. She had gone about a hundred and fifty metres when suddenly she stopped in mid-stride.

In the margin of his copy of Arithmetica, Pierre de Fermat had jotted the words I have a truly marvellous demonstration of this proposition which this margin is too narrow to contain.

The square had been converted to a cube, (x^3+y^3=z^3), and mathematicians had spent centuries looking for the answer to Fermat’s riddle. By the time Andrew Wiles solved the puzzle in the 1990s, he had been at it for ten years using the world’s most advanced computer programme.

And all of a sudden she understood. The answer was so disarmingly simple. A game with numbers that lined up and then fell into place in a simple formula that was most similar to a rebus.

Fermat had no computer, of course, and Wiles’s solution was based on mathematics that had not been invented when Fermat formulated his theorem. Fermat would never have been able to produce the proof that Wiles had presented. Fermat’s solution was quite different.

She was so stunned that she had to sit down on a tree stump. She gazed straight ahead as she checked the equation.

So that’s what he meant. No wonder mathematicians were tearing out their hair.

Then she giggled.

A philosopher would have had a better chance of solving this riddle.

She wished she could have known Fermat.

He was a cocky devil.

After a while she stood up and continued her approach through the trees. She kept the barn between her and the house.

[Peter van Velzen]

In the meantime I was acting foolishly because i forgot that there is not connection between the last digit in on base and the last digit in antother. The decimal nummber of …0 i may mean an end digit of 3 (10) or of 6(20) or of 2(30) or of 5 (40) or of (50) or of 4 (60) or of 0 (70) n base 7!
I am really stupid at times.

The probability of a solution would always have been small. Even for third powers one would have to be lucky to find For that situaties there are z-1 (say 99) candidates for making a solution with z being the solution, 20 which give a thirs power greater than half a million and 79 that are smaller. Combining one set with the other gives you a probability of less than 1600 in 1 for finding a specific number, while their sum is spanning a range between 500.001 and 1.493.039, giving a chance of less 1 in 600 for hitting the target. The problem is with increasing z, the chances don’t get any bigger! Going to z = 60.000 for instance will not get you any closer than. I compute 100 times 59,900 against 21.600 billon (usa style). which is about 1 in 35 million. Having 600 times the number of z’s to try and solve will not help much i fear. The chances are only dropping by a factor greater than 50.000 instead.

I did make some small discoveries. The difference between the greatest and the smallest variable (say z-x) must be divisible by a factor of the remaining one (say y) and probably so than the number of times it goes into z-x plus the number of times I goes into n, should be n or a multiple of n. The same is true for the difference between the two highest variables (say z-y) and the smallest one (say x) but i couldn’t rule out that being 1, so that might means nothing.

I suspect that the factor should at least be divisible by n once also, but I have thusfar failed to prove any of this. I was able to prove n cannot divide by four (unless I screwed up again, but I think this time I didn’t). So in 10 years in the 5th power I will probably have mastered this- Don’t tell me, to just go and read the official proof, for i am quite sure, I wouldn’t be able to understand it. . .

[Peter van Velzen]

So that’s what he meant. No wonder mathematicians were tearing out their hair.

Then she giggled.

A philosopher would have had a better chance of solving this riddle.

She wished she could have known Fermat.

He was a cocky devil.

After a while she stood up and continued her approach through the trees. She kept the barn between her and the house.

Beautifull story really, but after I gave up my wild goose chase on enddigits and concentrated on prime numbers in Pascal’s triangle. I actually think I have found a prove, that might have been something, Fermat had in mind. Everyone suspects there to be a flaw in it, but with my math skills surely: If Fermat did not see the flaw right away how could i?

Meanwhile it is only 5 pages long and took me no more than a day or two. So if this is what he meant, it is really ridiculously simple and it is really worth a giggle! Alas I can’t write subscripts or superscripts on the forum so I can’t post it here. But it is really silly. The prove for powers containing odd primes is especially stupid, and the difficult part was proving the fourth-power couldn’t work, Which I couldn’t have done it if I had not watched a video of Norman Wildberger on the parameterization of the Pythagoras triples.

For now I have saved my work and decided to sleep on it.
We will see if I am still giggling tomorrow. . .

[pallieter]

Briljant Peter, briljant!

Maar ga alstublieft door.

[Peter van Velzen]

Briljant Peter, briljant!

Maar ga alstublieft door.

May I remind you that this is the English section?

My (would-be) prove I thank to some inspiration of the numberphile video :"fool prove test for primes": http://www.youtube.com/watch?v=HvMSRWTE ... p3uz647V5A" onclick="window.open(this.href);return false; I But except for some information from this video by njwildberger: http://www.youtube.com/watch?v=OqxYLyGL ... 16&index=2" onclick="window.open(this.href);return false; only requires some high school algebra (which is all I am capable of anyway, never even managing an A in that field - I did manage an occasional B).
I wonder if someone can find a flaw with it. Everybody supposes the solution Fermat was thinking of was probably flawed, but with my limited skills I can – of course – not find one. So his solution might have been the same as mine.

[pallieter]

Shiny Peter, shiny.

But please walk further.