To begin, a model for squared numbers will be introduced and used to devise a method to create all Pythagorean (x^2 + y^2 = z^2) relationships. Equations will be derived from this process which indicate the existence of a Pythagorean equation in the model for squared numbers.
A model for higher powers of "n" will then be introduced. This model will be an extension of the model for squared numbers. Simple manipulations of this model will show that the "end game" packaging of quantities postulated to be x^n and y^n into spaces known to be x^n and yn requires that x, y, and z form a Pythagorean equation ! This is totally incompatible with the postulation that x^n + y^n = z^n where n >2.
The proof is thus Reductio ad Absurdum. . The recogniton of the afore-mentioned equations in the packaging process is the essence of the proof.
[zz)] [zz)] [zz)]Originally posted by matt grime
But your confidence appears to be misplaced. As I said it contains two obviously false assertions about generating pythagorean triples and claiming a reductio ad absurdum proof. The proof in case n=3 is half reasoned and not convincing and appears to make some unjustified conclusions about possible cases. That isn't to say the n=3 argument is wrong, just that it needs more examing. If you are genuine and not just a troll you should look at it and think about it - it uses nothing more than high school maths.
As the number of prime factors in "r" increases, the number of P. triplets doubles with each additional prime as follows:
"r" in prime factors Numerical value of "r" No. of P. triplets generated
2(1*3*5*7*11) 2310 16
2(1*3*5*7*11*13) 30030 32
2(1*3*5*7*11*13*17) 510510 64
2(1*3*5*7*11*13*17*19) 9699690 128
This method generates all valid Pythagorean triplets. Since any given P. triplet can be shown to have an even root number, "r", as shown in Figures 2 and 3, the P. triplet will eventually be generated as sequential even root numbers are selected and processed. The only limitations will be the capacity of the computational devices used.
Originally posted by matt grime
please read the first reply i gave on this where i asked if the generation was of the form I rewrote without the ambibuigty in the decompositions into prime factors (P_1P_2.... is used as is p_1p_2p_3p_4... and p_3p_4... which is ambiguous as it is not clear what the ...means) where I prove that 5,12,13 is not of the form specified.
Here you are deviding an odd number by a power of two. As a result, you are no longer dealing with whole numbers.Originally posted by Sariaht
(oa/2x)n +/- (ob/2x)n = ocn
well two odd numbers cannot become a third odd number.Originally posted by matt grime
yes, that is rather obviously what you must prove. but you've not done so.
No, but now we are dealing with binomials instead!Originally posted by suyver
Here you are deviding an odd number by a power of two. As a result, you are no longer dealing with whole numbers.
I guess I am just dumb. Binomial coefficients are of the formOriginally posted by Sariaht
No, but now we are dealing with binomials instead!
because x and y are uneven, and uneven numbers can be written 2n - 1.Originally posted by matt grime
what are you blathering about, sariaht?
if p^n+q^n=r^n then we may assuming that after dividing out by the largest power of 2 as necessary that at most one of these numbers is even. yes we know this. that is why we often look for p,q,r coprime - if any two of p,q and r are divisible by a number so is the third, therefore we may divide through and reduce to the case of coprime numbers, in particular at most one can be even, at most one is a multiple of 3 etc.
your reduction in the complexity of the problems doesn't reduce to anything at all that we didn't know already.
so why is it that expanding
(x/2+y/2)^n tells you anything?
bearing in mind you're now in the ring Z[1/2]
I didn't say it should. if g (sorry about the n) is high, then 1/2xg won't transform into a hole number whith the rest of the terms in the binomial funktion(s) ( if I'm not misstakeing ).Originally posted by matt grime
you have n meaning two different things. and i still fail to see why that tells you anything. in particular you've n ot excluded the case n=2 (n the exponenet) by any genuine means (ie said something non-trivial that is only true for every n>2)
there is no reason why 2^x should divide the 2m in 2m-1 either and even if it did, so what?