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Re: Pierre Fermat ever stack rectangles? If so, then he likely had a proof of FLT #1469 Correcting Math
Posted:
Jan 27, 2014 5:55 AM


Wow, never realised factorisation could be this simple.
So, please, run your "reasoning" past us for, say, 3 and 5 with odd powers:
f(2n+1) = 3^(2n+1) + 5^(2n+1), where n is natural (i.e., positive integer).
Here, I've calculated the first five for you, and even primefactored them.
f(3) = 152 = 2^3 * 19 f(5) = 3368 = 2^3 * 421 f(7)= 80312 = 2^3 * 10039 f(9) = 1972808 = 2^3 * 19 * 12979 f(11) = 49005272 = 2^3 * 23 * 266333
2^3 will be a factor of all the other ones too, and 2^4 etc. won't, but I guess you knew that, and why.
Here's your needed starts for the "exp"s you didn't list:
exp7 = 1 128 2187 16384 78125 279936 823543 2097152 4782969 10000000 19487171 35831808
exp9 = 1 512 19683 262144 1953125 10077696 40353607 134217728 387420489 1000000000 2357947691 5159780352
exp11 = 1 2048 177147 4194304 48828125 362797056 1977326743 8589934592 31381059609 100000000000
Off you go!
"Archimedes Plutonium" <plutonium.archimedes@gmail.com> wrote... Pierre Fermat ever stack rectangles? If so, then he likely had a proof of FLT #1469 Correcting Math
So, I proved FLT by simply noting that stacking two rectangles A and B, where both have one side equal to one another, and if a third rectangle exists in that exponent solution space which has a same rectangle dimensions of A + B, either proves by counterexample or proves no solution. A really simple proof.
So the question is, whether the margin of "Arithmetica" where Fermat wrote he had a proof of FLT, the question is, did Fermat ever engage in stacking rectangles in his mathematics work? Does the Arithmetica have a rectangle stacking section? Does any of Fermat's labors in mathematics have a stacking of rectangles or the decomposing of compound numbers into rectangles?
If Fermat had spent time on stacking rectangles, then it is likely he had a proof of FLT that he wrote in the margin. Only it would have been so simple of a proof, that he would have not deemed it important to write down.
Here is my proof of FLT:
______________________________________ Detailed Proof of FLT using condensedrectangles ______________________________________
It is a construction proof method for we show that it is impossible to construct A+B = C inside of a specific exponent.
Fermat's Last Theorem FLT conjecture says there are no solutions to the equation a^y + b^y = c^y where a,b,c,y are positive integers and y is greater than 2.
The number Space that governs FLT is this:
exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .}
exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .}
exp5 {1, 32, 243, 1024, 3125, 7776, 16807, 32,768, 59,049, 100,000, 161,051, 248,832, 371,293, . .}
exp6 ..... . . . .
So in FLT we ask whether there are any triples, A,B,C in any one of those _specific exponents_ such that A+B=C. In FLT, our solution space is only one particular exponent such as 3 or 4, or 5 to hunt down and find a A,B,C to satisfy A+B=C.
In the proof we use CondensedRectangles which is defined as a rectangle composed of unit squares of the cofactors of a number, except for 1 x number itself. So the number 27 in exp3 has Condensed Rectangles of 3x9 only. The number 125 in exp3 has condensed rectangles of 5x25 only, and the number 81 in exp4 has condensed rectangles of 3x27 and 9x9.
Now in the proof of FLT I am going to focus on two particular A and B where I have two condensed rectangles that share an Equal Side and where the A and B are of the same exponent:
2^3 + 2^3 = (2^4)
3^3 + 6^3 = (3^5)
For as you can easily see in exp3 we have the A, and B of 8 and 8 which is condensed rectangles of 2x4 and we can stack either on the 2 side or the 4 side in the one equation. And the 27 and 216 in the second equation, which are two condensed rectangles of 3x9 and 24x9 which we stack on the 9 side. But, now the question becomes, do we have a 16 condensed rectangle for the C in that of 2^3 + 2^3. Similarly, do we have a 243 condensed rectangle for that of a C in 3^3 + 6^3?
So, here we have the proof of FLT using condensedrectangles, for what we must show is that we can have a A and B but not all three of the A,B,C in the same exponent. Why is that? Because the demand of the A, B, C for A + B = C and having the same exponent will always deny the C to exist in that same exponent.
2^3 + 2^3 = 2^4
gets its 16 from out of exp3 because the next even number after 2 in exp3 is 4 and 4^3 places it as 4 times too large to satisfy A+B=C.
In the case of 3^3 + 6^3 = 3^5 we need to get a 243 out of exp3 but the demand of the next larger number than 6 with a 3 factor is 9 so that we would need a 9^3 to get us a 243 but 9^3 gets us 729 which is 3 times larger than 243.
So when confined to a single exponent, our C we need is going to be larger by at least a factor of 2 if even and a factor of 3 if odd.
So, in all cases of A,B,C where we have a A+B as condensed rectangles we cannot achieve a C, due to the fact that whether the C is even it is going to be a factor of at least 2 larger than the required C, or if C is odd it is going to be at least a factor of 3 larger than what is needed for a C. The only solutions of A+B=C is when we are allowed different exponents, but when confined to a single exponent such as 3, we cannot have a A+B=C, hence FLT.
Fermat's Last Theorem FLT is very easy to prove once you note that you need a different exponent for one of the A, B, C. That is about the only difference between proving FLT by condensed rectangles and proving Beal by condensed rectangles.
QED

