Re: [math-fun] Mystery tables
Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
Keith F. Lynch wrote:
Most such values of x are transcendental numbers. But not all. What rational values can x have? Can it have an irrational algebraic value?
(This is values of x where m^x - 2n^x = 1, with m,n positive integers.)
Right, i.e. where n is the xth power mean of 1 and m. My first table.
I betcha x is never an irrational algebraic number, and I betcha there will not in our lifetimes be sufficient mathematical technology to prove this.
Maybe so, but I've made some progress on rational values of x. It's 0 whenever m is the square of n, 1 whenever m+1 is 2n, and 2 whenever m^2+1 = 2n^2, i.e. where m and n are corresponding elements of A002315 and A001653. For instance 7,5 and 41,29. There are infinitely many solutions for each of x=0, 1, and 2. It would also be m^x+1 = 2n^x for other integers x, but according to http://www.math.mcgill.ca/darmon/pub/Articles/Research/18.Merel/paper.pdf (equation 1) there are no such integers. It's 1/j whenever there are integers a and j such that m = a^j and n = ((a+1)/2)^j. For instance 9,4 for 1/2 and 27,8 for 1/3. There are infinitely many solutions for every positive integer j. It's 2/j whenever j is odd and m and n are corresponding jth powers of A002315 and A001653, for instance 343,125, i.e. 7^3, 5^3. There are infinitely many solutions for every positive integer j. (For even j, the solution for 2/j is that for 1/(j/2).) I can't find any other rational values of x. Can you?
Hence my second table. I assume they're all transcendental except of course where x=y, in which case z=x=y. Can anyone prove this? Or find a counterexample?
(This is values of x where m^x + n^x = 2x^x, with m,n positive integers.)
I betcha x is never algebraic unless m=n, and I betcha there will not in our lifetimes be sufficient mathematical technology to prove this.
If they are irrational, they are always transcendental. See the Gelfond-Schneider theorem, also known as Hilbert's 7th problem. That leaves the remote possibility that there's some non-integer rational x.
On 08/02/2020 21:50, Keith F. Lynch wrote: [me:]
(This is values of x where m^x - 2n^x = 1, with m,n positive integers.)
[Keith:]
Right, i.e. where n is the xth power mean of 1 and m. My first table.
Yup. (I was just clarifying for the benefit of other readers, if any there be.)
Maybe so, but I've made some progress on rational values of x. It's 0 whenever m is the square of n, 1 whenever m+1 is 2n, and 2 whenever m^2+1 = 2n^2 [...]
It would also be m^x+1 = 2n^x for other integers x, but according to http://www.math.mcgill.ca/darmon/pub/Articles/Research/18.Merel/paper.pdf (equation 1) there are no such integers.
It's 1/j whenever there are integers a and j such that m = a^j and n = ((a+1)/2)^j. For instance 9,4 for 1/2 and 27,8 for 1/3. There are infinitely many solutions for every positive integer j.
It's 2/j whenever j is odd and m and n are corresponding jth powers of A002315 and A001653, for instance 343,125, i.e. 7^3, 5^3. There are infinitely many solutions for every positive integer j. (For even j, the solution for 2/j is that for 1/(j/2).)
I can't find any other rational values of x. Can you?
If m^(p/q) - 2n^(p/q) = 1 then either m^1/q and n^1/q are rational (hence integral) or not. If they _are_ rational then we are just looking at a solution of M^p-2N^p=1 and taking m=M^q, n=N^q. In that case, if what you said about integer values of x is right then there aren't going to be any solutions with p>2. So what if they _aren't_ rational? There's a theorem of Besicovitch cited here https://math.stackexchange.com/questions/150141/linear-independence-of-roots... one special case of which says: if each of m,n is divisible by some prime not dividing the other, then if P(m^1/q, n^1/q) = 0 with P a polynomial with rational coefficients, then either P has degree at least q in at least one variable, or else P is identically 0. If p = kq+r with 0 <= r < q, then m^(p/q) - 2n^(p/q) = 1 is the same as m^k m^(1/q)^r - 2n^k n^(1/q)^r = 1 which is a rational polynomial in m^1/q and n^1/q with both degrees < q, and it's not identically zero unless r=0, in which case m^1/q and n^1/q are rational. So in the irrational case the _other_ hypothesis of Besicovitch's theorem must fail, and one of m,n must be divisible by all the same primes as the other (plus perhaps some more). In that case, either one of m,n is 1 (impossible since then its 1/q power is 1, which is rational) or else they have a nontrivial common factor. Let s be a prime dividing both. It feels as if this should be impossible for mod-s-ish reasons but I'm not seeing how to formalize that right now and maybe it's wrong.
(This is values of x where m^x + n^x = 2x^x, with m,n positive integers.)
I betcha x is never algebraic unless m=n, and I betcha there will not in our lifetimes be sufficient mathematical technology to prove this.
If they are irrational, they are always transcendental. See the Gelfond-Schneider theorem, also known as Hilbert's 7th problem. That leaves the remote possibility that there's some non-integer rational x.
How do you get this from Gelfond-Schneider? Sorry if I'm just being dim, but while this is clearly G-S-esque I don't see how G-S is enough to prove it. -- g
participants (2)
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Gareth McCaughan -
Keith F. Lynch