Rupert wrote: > email@example.com wrote: > > The widespread usage of flawed ideas in number theory has put me in the > > position of not just proving the errors and getting publication of > > proof in a peer reviewed math journal, but necesitated the finding of a > > second proof as well due to extraordinary resistance to such a shocking > > result. > > > > I have finally found the long sought after second proof. > > > > Start with > > > > P(x) = 175x^2 - 15x + 2 > > > > and do your analysis on the factorization > > > > P(x) = (f(x) + 2)*(g(x) + 1) > > > > where f(0) = g(0) = 0, where the functions are otherwise to be > > determined. > > > > So I have > > > > 175x^2 - 15x + 2 = (f(x) + 2)*(g(x) + 1) > > > > and next I do some simple algebra starting with multiplying both sides > > by 7: > > > > 7*(175x^2 - 15x + 2) = 7*(f(x) + 2)*(g(x) + 1) > > > > now I re-order in a special way on the left side and pick one way to > > multiply by 7 on the right: > > > > (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2 = (f(x) + 2)*(7g(x) + 7) > > > > That work is done so that you have 5 and 7 on the left like a > > polynomial with variables, except of course 5 and 7 are not variable, > > but the operations are valid to this point. > > > > But now I need to balance the right side so let > > > > f(x) = 5a_1(x) + 5 > > > > and > > > > 7g(x) =5a_2(x) > > > > which gives > > > > (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 = (5a_1(x) + 7)*(5a_2(x) + 7) > > > > allowing me to find a solution of the a's where they are roots of > > > > a^2 - (7x-1)a + (49x^2 - 14x) = 0. > > > > The problem though is in getting to this point--using some VERY simple > > algebra--I multiplied g(x) times 7 to get 5a_2(x), so it makes sense > > that a_2(x) has 7 as a factor. > > > > Only if you have some reason to think that g(x) is an algebraic > integer. You have no reason to think that whatsoever. > > > BUT if you plug in some numbers like let x=1, you have > > > > a^2 - 6a + 35 = 0 > > > > which solves as > > > > a = (6 +/- sqrt(-104))/2 > > > > and the result is that one of the roots has 7 as a factor. > > > > But provably neither do in the ring of algebraic integers. > > > > In arguing against this result posters have claimed that neither root > > has 7 itself as a factor, but instead each root shares factors in > > common with 7, but from the start with > > > > 7g(x) =5a_2(x) > > > > if a_2(x) does NOT have 7 itself as a factor then g(x) must divide off > > some factors in common with 7. > > > > But going back to > > > > 175x^2 - 15x + 2 = (f(x) + 2)*(g(x) + 1) > > > > note that 175x^2 - 15x + 2 is always even when x is an integer. > > > > But now let > > > > f(x) = 2*g'(x) and g(x) = f'(x)/2 > > > > and making the substitutions and simplifying gives > > > > 175x^2 - 15x + 2 = (g'(x) + 1)*(f'(x) + 2) > > > > and the same approach as before gives the same solution for the a's, > > No, it does not. > > We have > > f(x)=5a_1(x)+5 > g(x)=5a_2(x)/7 > > where the a's are roots of > > a^2-(7x-1)a+(49x^2-14x)=0. > > Now, if we let g'(x)=f(x)/2 and f'(x)=2g(x), we have > > f'(x)=2(5a_2(x)/7)=5b_1(x)+5 where b_1(x)=2(a_2(x)/7)-1 > g'(x)=(5a_1(x)+5)/2=5b_2(x)/7 where b_2(x)=7(a_1(x)+1)/2. > > We have a_1(x)+a_2(x)=7x-1, a_1(x)a_2(x)=49x^2-14x. > > By working out b_1(x)+b_2(x) and b_1(x)b_2(x), we can get an equation > of which the b's will be roots. It will not be the same equation. >
My apologies, this is not the way to get the minimum polynomial of the b's. The two b's will have different minimum polynomials.
b_1(x) will be a root of
b_2(x) will be a root of
As I say, I assume this is what you are trying to do, namely you are letting g'(x)=f(x)/2 and f'(x)=2g(x), and then you are defining b's so that
If that is what you are doing, then the b's will not be roots of the same equation as the a's.
If you are doing something else, you'd better explain what. > > where now > > > > 7g'(x) = 7*f(x)/2 = 5a_2(x) > > > > and f(x) would have to divide out factors from 7, so there would be a > > contradiction. > > > > Therefore the claim that g(x) can simply divide off factors from 7, so > > that 7*g(x) does not have 7 as a factor is refuted. > > > > Note this claim is an objection raised to protect the current role of > > the ring of algebraic integers as provably 7*g(x) does NOT have 7 as a > > factor in that ring, with non-zero integer x as noted above. > > > > So what is the resolution to this weirdness? > > > > Well, consider evens and notice that 2 is not a factor of 6 with only > > evens because 3 is not even. > > > > So it's not very complicated since the problem with the ring of > > algebraic integers is about exclusion. > > > > Some numbers are excluded because they are not root of monic > > polynomials with integer coefficients, so you can get weird stuff, and > > an analogy is with evens where odds are excluded, so 2 is not a factor > > of 6 with only evens. > > > > Proving the odd behavior of the ring of algebraic integer is easy. I > > just did it yet again above. > > > > Explaining it simply is easy as well, as I just did, yet again with the > > 2 and 6 analogy. > > > > But there are a lot of people around the world who learned mathematical > > ideas that have been shown to be flawed and so far that group has > > resisted publication, bizarre retraction of these ideas, and the death > > of the math journal--to keep using the wrong ideas, and teach them to > > others. > > > > While posters have routinely lied about the details of the argument and > > objected for years in a steady attempt to deny what is mathematically > > correct. > > > > And meanwhile undergraduates have been taught mathematical ideas proven > > wrong years ago. > > > > I do not know what else to do here. If math people will not follow > > mathematical proof, and even second proof, then what can anyone do? > > > > If the will for error is too strong for rational argument to work, then > > it seems there are few options in this situation except to grieve over > > the fall of humanity and the failure of civilization in this area. > > > > > > James Harris