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"Fermat's Last Theorem" by Amir D. Aczel

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Fermat's Last Theorem

Unlocking the Secret of an Ancient Mathematical Problem.

Amir D. Aczel

 

Your knowledge of math won't necessarily be challenged by this short 137 page hardcover edition. It presents a look at the mathematicians who provided various keys to Professor Andrew Wiles 1993 confirmation of a note made in a margin three centuries earlier.

 

Pierre de Fermat had written around 1637 inthe "Arithmetica" written by the Greek mathematician Diophantus, next to a problem on breaking down a squared number into two squares:

 

On the other hand, it is impossible to separate a cube into two cubes, or a biquadrate into two biquadrats, or generally any power except a square into two powers with the same exponent. I have discovered a truly marvelous proof of thes, which, however, the margin is not large enough to contain.

 

Fermat broke ground developing the main ideas of calculus 13 years before Newton was born.

Amir suggests that the origins of A2+B2=C2 being resolved for integers go back to at least Circa 2000 B.C. with connections to wealth, land surveying, and religion. He spends about 1/3 of the book various observations and methods that were developed and available to Fermet.

 

The book is filled with thumbnail biographies of key mathematicians where available, and what their contributions were - speckled with a few mathematical representations, all taking place after Fermet, and viewed as instrumental for Andrew Wiles to have provided the presentation he did. Amir closes the book withthe statement:

 

So whether or not Fermat did possess a "truly marvelous proof" of his theorem, one that could not fit in the margin of his book, will forever remain his secret.

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Given what it took to finally prove FLT  is is fairly clear that Fermat did NOT have a proof that would stand up to modern standards of rigor.

 

If you want to see how Wiles did it look at 

http://www.cs.berkeley.edu/~anindya/fermat.pdf

 

It is only 109 pages long.  Fermat claimed his proof fit on the margin of a text book.  

 

ruveyn1

Edited by ruveyn1

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Sitting down with a peice of paper after finishing the book, I soon found it filled with numbers. Let me share a few with you.

 

Let's start with the squares.

 

12=1, 22=4, 32=9, 42=16, 52=25, 62=36, 72=49, 82=64, 92=81, 102=100, 112=121, 122=144, 132=169, 142=196, 152=225, 162=256, 172=289, 182=324, 192=361, 202=400

 

Moving onto the cubes we find,

 

13=1, 23=8, 33=27, 43=64, 53=125, 63=216, 73=343, 83=512, 93=729, 103=1000, 113=1331, 123=1728, 133=2197, 143=2744, 153=3375, 163=4096, 173=4913, 183=5832, 193=6859, 203=8000
 

Raising to the power of 4 yields,

 

14=1, 24=16, 34=81, 44=256, 54=625, 64=1296, 74=2401, 84=4096, 94=6561, 104=10000, 114=14641, 124=20736, 134=28561, 144=38416, 154=50625, 164=65536, 174=83521, 184=104976, 194=130321, 204=160000


Looking to the power of 5ths,

 

15=1, 25=32, 35=243, 45=1024, 55=3125, 65=7776, 75=16807, 85=32768, 95=59049, 105=100000, 115=161051, 125=248832, 135=371293, 145=537824, 155=759375, 165=1048576, 175=1419857, 185=1889568, 195=2476099, 205=3200000

 

And lastly at the power of 6ths.

 

16=1, 26=64, 36=729, 46=4096, 56=15625, 66=46656, 76=117649, 86=262144, 96=531441, 106=1000000, 116=1771561, 126=2985984, 136=4826809, 146=7529536, 156=11390625, 166=16777216, 176=24137569, 186=34012224, 196=47045881, 206=64000000

 

This does give quite a array of numbers.

 

==================================================================================================================================

 

Let's take this a step further.

 

Starting with the squared differences,

4-1=3, 9-4=5, 16-9=7, 25-16=9, 36-25=11, 49-36=13, 64-49=15, 81-64=17, 100-81=19, 121-100=21, 144-121=23, 169-144=25, 196-169=27, 225-196=29, 256-225=31, 289-256=33, 324-289=35, 361-324=37, 400-361=39

 

The cubed differences are as follows,

8-1=7, 27-8=19, 64-27=37, 125-64=61, 216-125=91, 343-216=127, 512-343=169, 729-512=217, 1000-729=271, 1331-1000=331, 1728-1331=397, 2197-1728=469, 2744-2197=547, 3375-2744=631, 4096-3375=721, 4913-4096=817, 5832-4913=919, 6859-5832=1027, 8000-6859=1141

 

The differences raised to the 4th power are,

16-1=15, 81-16=65, 256-81=175, 625-256=369, 1296-625=671, 2401-1296=1105, 4096-2401=1695, 6561-4096=2465, 10000-6561=3439, 14641-10000=4641, 20736-14641=6095, 28561-20736=7825, 38416-28561=9855, 50625-38416=12209, 65536-50625=14911, 83521-65536=17985, 104976-83521=21455, 130321-104976=25345, 160000-130321=29679
 

The differences raised to the 5th power are,

32-1=31, 243-32=211, 1024-243=781, 3125-1024=2101, 7776-3125=4651, 16807-7776=9031, 32768-16807=15961, 59049-32768=26281, 100000-59049=40951, 161051-100000=61051, 248832-161051=87781, 371293-248832=122461, 537824-371293=166531, 759375-537824=221551, 1048576-759375=289201, 1419857-1048576=371281, 1889568-1419857=469711, 2476099-1889568=586531, 3200000-2476099=723901

 

And lastly, the differences raised to the 6th power are,

64-1=63, 729-64=665, 4096-729=3367, 15625-4096=11529, 46656-15625=31031, 117649-46656=70993, 262144-117649=144495, 531441-262144=269297, 1000000-531441=468559, 1771561-1000000=771561, 2985984-1771561=1214423, 4826809-2985984=1840825, 7529536-4826809=2702727, 11390625-7529536=3861089, 16777216-11390625=5386591, 24137569-16777216=7360353, 34012224-24137569=9874655, 47045881-34012224=13033657, 64000000-47045881=16954119

 

=========================================================================================================================================

 

Summarizing:

 

Squared:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39

Cubed:

1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141

Quadratic:

1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679

To the 5th,

1, 31, 211, 781, 2101, 4651, 9031, 15961, 26281, 40951, 61051, 87781, 122461, 166531, 221551, 289201, 371281, 469711, 586531, 723901

To the 6th,

1, 63, 665, 3367, 11529, 31031, 70993, 144495, 269297, 468559, 771561, 1214423, 1840825, 2702727, 3861089, 5386591, 7360353, 9874655, 13033657, 16954119

 

=========================================================================================================================================

 

Reviewing the squared sequence, the assigned difference can be represented as:2*X+1.

2*0+1=1, 2*1+1=3 2*2+1=5, 2*3+1=7, 2*4+`=9, 2=5+1=11, 2*6+1=13, 2*7+1=15, 2*8+1=17, 2*9+1=19, 2*10+1=21, etc.

 

What is not as obvious is for the cubed sequence: X2+(X+1)(2*X+1)

02+(0+1)(2*0+1)=1, 12+(1+1)(2*1+1)=7, 22+(1+1)(2*1+1)=7, 22+(2+1)(2*2+1)=19, 32+(3+1)(2*3+1)=37, 42+(4+1)(2*4+1)=61, etc.

 

Stepping to the next sequence, what becomes clearer is X3+((X+1)(X2+(X+1)(2*X+1)). This gives us:

03+(0+1)(02+(0+1)(2*0+1))=1, 13+(1+1)(12+(1+1)(2*1+1))=15, 23+(2+1)(22+(2+1)(2*2+1))=65, 33+(3+1)(32+(3+1)(2*3+1))=175, etc.

 

Extrapolating to the 5th power: X4+(X+1)((X3+(X+1)((X2+(X+1)(2*X+1))) yields:

04+(0+1)(03+(0+1)((02+(0+1)(2*0+1)))=1, 14+(1+1)(13+(1+1)((12+(x1+1)(2*1+1))=31, 24+(2+1)(23+(2+1)((22+(2+1)(2*2+1)))=211, 34+(3+1)(33+(3+1)((32+(3+1)(2*3+1)))=671, etc.

 

And for this example, taking it finally to the 6th power: X5+(X+1)((X4+(X+1)((X3+(X+1)((X2+(X+1)(2*X+1))))

05+(0+1)(04+(0+1)(03+(0+1)(02+(0+1)(2*0+1))))=1, 15+(1+1)(14+(1+1)(13+(1+1)(12+(1+1)(2*1+1))))=63, 25+(2+1)(24+(2+1)(23+(2+1)(X2+(2+1)(2*2+1))))=665, 35+(3+1)(34+(3+1)(33+(3+1)(32+(3+1)(2*3+1))))=3367, etc.

 

Could Fermat have induced from a pattern such as this, that for a power greater than 2, An+Bn=Cn could not have been resolved into an integral relationship?

 

 

fixed 761 from 671 on the 5th power 34,, & parenthesis fixed on various others.

Edited by dream_weaver

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Sitting down with a peice of paper after finishing the book, I soon found it filled with numbers. Let me share a few with you.

 

Let's start with the squares.

 

12=1, 22=4, 32=9, 42=16, 52=25, 62=36, 72=49, 82=64, 92=81, 102=100, 112=121, 122=144, 132=169, 142=196, 152=225, 162=256, 172=289, 182=324, 192=361, 202=400

 

Moving onto the cubes we find,

 

13=1, 23=8, 33=27, 43=64, 53=125, 63=216, 73=343, 83=512, 93=729, 103=1000, 113=1331, 123=1728, 133=2197, 143=2744, 153=3375, 163=4096, 173=4913, 183=5832, 193=6859, 203=8000

 

Raising to the power of 4 yields,

 

14=1, 24=16, 34=81, 44=256, 54=625, 64=1296, 74=2401, 84=4096, 94=6561, 104=10000, 114=14641, 124=20736, 134=28561, 144=38416, 154=50625, 164=65536, 174=83521, 184=104976, 194=130321, 204=160000

Looking to the power of 5ths,

 

15=1, 25=32, 35=243, 45=1024, 55=3125, 65=7776, 75=16807, 85=32768, 95=59049, 105=100000, 115=161051, 125=248832, 135=371293, 145=537824, 155=759375, 165=1048576, 175=1419857, 185=1889568, 195=2476099, 205=3200000

 

And lastly at the power of 6ths.

 

16=1, 26=64, 36=729, 46=4096, 56=15625, 66=46656, 76=117649, 86=262144, 96=531441, 106=1000000, 116=1771561, 126=2985984, 136=4826809, 146=7529536, 156=11390625, 166=16777216, 176=24137569, 186=34012224, 196=47045881, 206=64000000

 

This does give quite a array of numbers.

 

==================================================================================================================================

 

Let's take this a step further.

 

Starting with the squared differences,

4-1=3, 9-4=5, 16-9=7, 25-16=9, 36-25=11, 49-36=13, 64-49=15, 81-64=17, 100-81=19, 121-100=21, 144-121=23, 169-144=25, 196-169=27, 225-196=29, 256-225=31, 289-256=33, 324-289=35, 361-324=37, 400-361=39

 

The cubed differences are as follows,

8-1=7, 27-8=19, 64-27=37, 125-64=61, 216-125=91, 343-216=127, 512-343=169, 729-512=217, 1000-729=271, 1331-1000=331, 1728-1331=397, 2197-1728=469, 2744-2197=547, 3375-2744=631, 4096-3375=721, 4913-4096=817, 5832-4913=919, 6859-5832=1027, 8000-6859=1141

 

The differences raised to the 4th power are,

16-1=15, 81-16=65, 256-81=175, 625-256=369, 1296-625=671, 2401-1296=1105, 4096-2401=1695, 6561-4096=2465, 10000-6561=3439, 14641-10000=4641, 20736-14641=6095, 28561-20736=7825, 38416-28561=9855, 50625-38416=12209, 65536-50625=14911, 83521-65536=17985, 104976-83521=21455, 130321-104976=25345, 160000-130321=29679

 

The differences raised to the 5th power are,

32-1=31, 243-32=211, 1024-243=781, 3125-1024=2101, 7776-3125=4651, 16807-7776=9031, 32768-16807=15961, 59049-32768=26281, 100000-59049=40951, 161051-100000=61051, 248832-161051=87781, 371293-248832=122461, 537824-371293=166531, 759375-537824=221551, 1048576-759375=289201, 1419857-1048576=371281, 1889568-1419857=469711, 2476099-1889568=586531, 3200000-2476099=723901

 

And lastly, the differences raised to the 6th power are,

64-1=63, 729-64=665, 4096-729=3367, 15625-4096=11529, 46656-15625=31031, 117649-46656=70993, 262144-117649=144495, 531441-262144=269297, 1000000-531441=468559, 1771561-1000000=771561, 2985984-1771561=1214423, 4826809-2985984=1840825, 7529536-4826809=2702727, 11390625-7529536=3861089, 16777216-11390625=5386591, 24137569-16777216=7360353, 34012224-24137569=9874655, 47045881-34012224=13033657, 64000000-47045881=16954119

 

=========================================================================================================================================

 

Summarizing:

 

Squared:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39

Cubed:

1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141

Quadratic:

1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679

To the 5th,

1, 31, 211, 781, 2101, 4651, 9031, 15961, 26281, 40951, 61051, 87781, 122461, 166531, 221551, 289201, 371281, 469711, 586531, 723901

To the 6th,

1, 63, 665, 3367, 11529, 31031, 70993, 144495, 269297, 468559, 771561, 1214423, 1840825, 2702727, 3861089, 5386591, 7360353, 9874655, 13033657, 16954119

 

=========================================================================================================================================

 

Reviewing the squared sequence, the assigned difference can be represented as:2*X+1.

2*0+1=1, 2*1+1=3 2*2+1=5, 2*3+1=7, 2*4+`=9, 2=5+1=11, 2*6+1=13, 2*7+1=15, 2*8+1=17, 2*9+1=19, 2*10+1=21, etc.

 

What is not as obvious is for the cubed sequence: X2+(X+1)(2*X+1)

02+(0+1)(2*0+1)=1, 12+(1+1)(2*1+1)=7, 22+(1+1)(2*1+1)=7, 22+(2+1)(2*2+1)=19, 32+(3+1)(2*3+1)=37, 42+(4+1)(2*4+1)=61, etc.

 

Stepping to the next sequence, what becomes clearer is X3+((X+1)(X2+(X+1)(2*X+1)). This gives us:

03+(0+1)(02+(0+1)(2*0+1))=1, 13+(1+1)(12+(1+1)(2*1+1))=15, 23+(2+1)(22+(2+1)(2*2+1))=65, 33+(3+1)(32+(3+1)(2*3+1))=175, etc.

 

Extrapolating to the 5th power: X4+(X+1)((X3+(X+1)((X2+(X+1)(2*X+1))) yields:

04+(0+1)(03+(0+1)((02+(0+1)(2*0+1)))=1, 14+(1+1)(13+(1+1)((12+(x1+1)(2*1+1))=31, 24+(2+1)(23+(2+1)((22+(2+1)(2*2+1)))=211, 34+(3+1)(33+(3+1)((32+(3+1)(2*3+1)))=671, etc.

 

And for this example, taking it finally to the 6th power: X5+(X+1)((X4+(X+1)((X3+(X+1)((X2+(X+1)(2*X+1))))

05+(0+1)(04+(0+1)(03+(0+1)(02+(0+1)(2*0+1))))=1, 15+(1+1)(14+(1+1)(13+(1+1)(12+(1+1)(2*1+1))))=63, 25+(2+1)(24+(2+1)(23+(2+1)(X2+(2+1)(2*2+1))))=665, 35+(3+1)(34+(3+1)(33+(3+1)(32+(3+1)(2*3+1))))=3367, etc.

 

Could Fermat have induced from a pattern such as this, that for a power greater than 2, An+Bn=Cn could not have been resolved into an integral relationship?

 

 

fixed 761 from 671 on the 5th power 34,, & parenthesis fixed on various others.

It does not follow.  Just because something fails for the first gazillion does not mean it will not succeed for gazillion plus 1. 

 

You cannot, in general disprove non existence of something just by exhibiting a finite number of failures. 

 

Mathematicians very rarely work by empirical induction (not to be confused with arithmetic induction which is something completely different).

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Using 2x+1

1, 2, 3,  4,   5,   6,   7,   8,   9,   10,   11,   12,   13,   14,   15,   16,   17,   18,   19,   20 <--- x

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400 <--- x2

  3, 5,  7,  9,  11, 13, 15, 17, 19,   21,   23,   25,   27,   29,   31,   33,   35,   37,   39 <--- 2x+1

 

Little was stated in the book about finding Pythagorian triples. Wikipedia and Wolfram have some formulas that can be used. I've not tried to contrast what I've done here with what has been done there at this point.

From the last row, the 9 appears between the 16 and 25 of the second row. 3,4,5

Next we find 25 between 144 and 169 yielding another triple of 5,12,13

To find what works with 7, 2*x+1=49, x would be between 24 and 25 yielding 7,24,25

Starting with 3,4,5 we also can see that 2(3,4,5) = 6,8,10 and 3(3,4,5) = 9,12,15 and 4(3,4,5) = 12,16,20

This can also be done with our earlier triplet using 2(5,12,13) to get 10,24,26

 

While certainly not exhaustive, there are some definite patterns available to be explored.

 

Moving on to x2+(x+1)(2x+1) we get the following:

1, 2,   3,   4,     5,     6,      7,     8,       9,     10,      11,      12,      13,      14,      15,      16,      17,      18,      19,       20 <--- x

1, 8,  27, 64, 125, 216,  343,  512,  729,  1000,  1331,  1728,  2197,  2744,  3375,  4096,  4913,  5832,  6859,   8000 <--- x3

  7, 19, 37, 61,   91,  127,  169,  217,  271,    331,    397,    469,    547,    631,    721,    817,    919,    1027,   1141 <--- x2+(x+1)(2x+1)

 

x3+(x+1)(x2+(x+1)(2x+1))

1,   2,    3,      4,        5,       6,          7,          8,          9,          10,         11,          12,         13,           14,            15,           16,            17,              18,            19,               20 <--- x

1,  16,  81,   256,   625,   1296,    2401,    4096,    6561,    10000,   14641,    20736,    28561,    38416,      50625,      65536,      83521,      104976,     130321,      160000 <---x4

  15,  65, 175,   369,   671,    1105,    1695,    2465,    3439,      4641,      6095,      7825,     9855,       12209,      14911,      17985,      21455,       25345,        29679 <--- x3+(x+1)(x2+(x+1)(2x+1))

 

Perhaps noting the number of prime numbers in the cubed series this may have inspired Fermat to develop his formula for primes, 22^n+1, which was discovered to be false.

The fact is, a pattern can be isolated in the differences, and it can be formulated along the lines of xn+(x+1)( . . . .(x2+(x+1)(2x+1))).

Arranging the results in a way that other patterns emerged for the squared series failed to provide similar patterns for myself for up to x8 for the first 20 integers before excel started converting the answers into scientific notation.

 

What Fermat described as a "truly marvelous proof" that would not fit in the margin where he made his notation, has not been discovered.

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What Fermat described as a "truly marvelous proof" that would not fit in the margin where he made his notation, has not been discovered.

It probably does not exist.

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Given what it took to finally prove FLT  is is fairly clear that Fermat did NOT have a proof that would stand up to modern standards of rigor.

 

If you want to see how Wiles did it look at 

http://www.cs.berkeley.edu/~anindya/fermat.pdf

 

It is only 109 pages long.  Fermat claimed his proof fit on the margin of a text book.  

 

ruveyn1

 

I have not read the proof cited above, but I consider it possible that Fermat might have had, or someone else might discover, a much more elegant and insightful proof that is much shorter than 109 pages. I'm sure it wouldn't be the first time that a long proof could be replaced by a much shorter one, but I don't have any examples at hand to cite.

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I have not read the proof cited above, but I consider it possible that Fermat might have had, or someone else might discover, a much more elegant and insightful proof that is much shorter than 109 pages. I'm sure it wouldn't be the first time that a long proof could be replaced by a much shorter one, but I don't have any examples at hand to cite.

I really doubt it.  Some of the greatest mathematicians of all time have had a go at it,  and Wiles was the one who finally did it after 300+ years.  I will believe there is a short elementary proof when I see it with my own eyes.l

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Considering that Aristotle may have induced his rules of logic methodology from pre-Euclidean geometry, the application of todays modern standards of rigor within the science of mathematics sometimes elude to questions I have difficulty discovering how to articulate. Cubes, Quads, Quints, etc, introduce methods of using number to measure number  differently than than the number line. articulates.

 

Being enumerated among the greatest mathematicians of all times does not ensure adherence to objectivity. While Wiles finally offered a proof that is acceptable to the mathematical community, does not ensure that it is indeed objective, in an Objective sense of the term.

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Fermat’s Last Theorem

 

Pythagrean Theorem states that the hypotenuse of a right triangle can be resolved by taking the square root of the sum of the sides.

 

Pythagrean triples have been discovered and used historically.

 

Fermat claimed in a margin of a lost book that there is no integer solution for An>2+Bn>2=Cn>2

 

I now agree with him, and not for the fact that a 109 page dissertation was written on the subject that I have not read.

 

The diagonal of a square involves the square root of 2. If a side is an integer, the diagonal is a factor involving the square root of 2. If the diagonal is an integer, the sides will have a factor involving the square root of 2.

 

To solve for the greater diagonal on a cube also invokes the square root of 2. The formula breaks down to square root of 3*x2. This can be reduced to the square root of 3, also an irrational number.

 

Consider a cube with a side of 1, the diagonal is the square root of 3.

Consider a cube with a side of 2, the diagonal is 2*square root of 3.

 

The formula works out to (x2+x2+x2)1/2

 

To end up with an integer for the answer, the side would have to be an irrational number.

 

The formula for exponents 3 or greater.

Square root of (x2+x2+x(x-2))

(For powers greater than 3, the result ends up with multiple cubes. 24 would be 2 cubes of 8 for a total of 16. 88 would be 32768 cubes of 4096)

 

It doesn’t matter whether there is 1 cube, or a gazillion cubes, the answer remains that a diagonal of a cube contains a factor of the square root of 3, or an integer was not used to derive the answer.

Edited by dream_weaver

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It doesn’t matter whether there is 1 cube, or a gazillion cubes, the answer remains that a diagonal of a cube contains a factor of the square root of 3, or an integer was not used to derive the answer.

The real task was showing that x^n + y^n = z^n  did not have nonzero solutions x, y, z  for ALL n > 2. 

 

That stymied the world's best mathematicians for over 300 years.

 

Here is a problem which if you can solve it will make you as famous as Dr. Wiles.

 

It is the proof of the Collatz conjecture.

 

I will define a function on the integers as follows:

 

T(n) = n/2  if n is even

T(n) = (3*n + 1)/2  if n is odd.

 

Here is the problem:  show that starting with any integer N  if you keep applying the function T  you will eventually get to 1.

 

Example.. N = 7

7, 11, 17, 26, 13, 20, 10, 5, 4, 2, 1

 

That problem has been open since around 1950.  Paul Erdos one of the great great mathematicians was of the opinion that mankind was not ready to solve that problem. 

 

It has already been show for N < 10^20  this is true,  but a proof for ALL N  remains to be found.

 

All attempts to prove that the general proposition is undecidable have also failed.

 

ruveyn1

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(x2+x2+xx-2)1/2 should show that xn+yn=zn has no nonzero solutions x,y,z for ALL n>2. I do not have the mathematical forte to convert it at this time. It is true, for the same reason that the hypotenuse of a square is incommensurate to its sides.

 

I have a bookmark for consideration of Collatz conjecture. If fame were my goal, . . . who knows.

 

By the way, 5-->(3*n+1/2)=8, not 4, although 8, 4, 2, 1 does finish the hypothesis.

Edited by dream_weaver

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(x2+x2+xx-2)1/2 should show that xn+yn=zn has no nonzero solutions x,y,z for ALL n>2. I do not have the mathematical forte to convert it at this time. It is true, for the same reason that the hypotenuse of a square is incommensurate to its sides.

 

I have a bookmark for consideration of Collatz conjecture. If fame were my goal, . . . who knows.

 

By the way, 5-->(3*n+1/2)=8, not 4, altsohough 8, 4, 2, 1 does finish the hypothesis.

sorry about the error.  A trivial corollary:  once a collatz sequence hits a power of two,  the game is over.

 

You say you have a proof :  "(x2+x2+xx-2)1/2 should show that xn+yn=zn has no nonzero solutions x,y,z for ALL n>2. I do not have the mathematical forte to convert it at this time. It is true, for the same reason that the hypotenuse of a square is incommensurate to its sides."

 

Claims are cheap.  Valid proofs are dear:   Show the proof  or be still. 

Edited by ruveyn1

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Alright. Now lets see if we can attach this thing. (Between here)

(here)post-7393-0-02377600-1362775985_thumb.jp(here)

Its hard to see (and here.)

 

 

Imagine walking into a building at the door. The building is X by X. Walk across the diagonal to the corner opposite the door. Climb the stairwell crossing back across the diagonal to the corner over the door. This is (3X2)^1/2

 

The next solution where I corrected my original formula, is for a 2 story, 3 story, 4 story etc. The stairwell starts in the same corner, only it climbs to the top corner over the door of the destination floor .

 

(X2+X2+((exponent-2)X)2)1/2

Edited by dream_weaver

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As to Collantz Conjecture that should work for any odd number.

 

T(n) = n/2  if n is even

T(n) = (3*n + 1)/2  if n is odd.

 

T(n) = n/2  if n is even

T(n) = (5*n + 1)/2  if n is odd.

 

T(n) = n/2  if n is even

T(n) = (7*n + 1)/2  if n is odd.

 

T(n) = n/2  if n is even

T(n) = (9*n + 1)/2  if n is odd.

...

Edited by dream_weaver

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As to Collantz Conjecture that should work for any odd number.

 

T(n) = n/2  if n is even

T(n) = (3*n + 1)/2  if n is odd.

 

T(n) = (5*n + 1)/2  if n is odd.

T(n) = (7*n + 1)/2  if n is odd.

T(n) = (9*n + 1)/2  if n is odd.

...

Prove that and you will become an Immortal.

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Alright. Now lets see if we can attach this thing.

attachicon.gifFermatsLastTheorem.jpg

 

Imagine walking into a building at the door. The building is X by X. Walk across the diagonal to the corner opposite the door. Climb the stairwell crossing back across the diagonal to the corner over the door. This is (3X2)^1/2

 

The next solution where I corrected my original formula, is for a 2 story, 3 story, 4 story etc. The stairwell starts in the same corner, only it climbs to the top corner over the door of the destination floor .

 

(X2+X2+((exponent-2)X)2)1/2

Do you have any idea of what constitutes a rigorous proof?

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Nope.

 

I had a little pre-calc 30 years ago.

 

But sooner or later, (3n+1)=2x, as does (5n+1) or (7n+1)

 

3n+1 does for 1, 5 and more,

5n+1 does for 3, 7 and more.

7n+1 does for 1, 5 and more.

 

I would tend to believe that is why it works.

Edited by dream_weaver

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Nope.

 

I had a little pre-calc 30 years ago.

Do two things.

 

1  Learn what a proof is

2. Write you proof out carefully and have a professional mathematician vet your proof for errors.

 

If you can prove FLT  by elementary means and you are under 45 years of age,  I guarantee you will get a Fields Medal.

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Already past that point. I just enjoy tinkering with this stuff occasionally. The drawing I produced is more in line with my forte. I have to envision what it looks like in order to create it.

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The more I look at my final formula using the exponent, I do not think it accurate represents the number of floors. I think there are many more, and are influenced by the number being exponented in some way..

Up to the (X2+X2+X2) is the means of deriving that figure. Like the pythagorean triple cannot occur in a square, the cube exhibits similar properties concretized in this form.

Edited by dream_weaver

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Another way of looking at the problem.

 

Take a cube of 10 X 10 X 10 boxes,

Take 8 X 8 X 8 boxes away,

Take 2 X 2 X 2 boxes away. 

What is left?

 

103-83-23 cannot be 0

The remainder is not enough to make 83 and 33 or 93 and 23.

 

(A+B )3-A3-B3>0

(A+1)3+B3>(A+B )3

A3+(B+1)3>(A+B )3

Edited by dream_weaver

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