[math-fun] To Allan Re: "I don't find this proof beautiful" quote?
Allan, I just asked essentially the identical question to math-fun a few weeks ago, under a subject line something like "Unsolved number theory problem". There were some interesting responses. —Dan -----Original Message-----
From: Allan Wechsler <acwacw@gmail.com> Sent: May 30, 2018 3:04 PM To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] "I don't find this proof beautiful" quote?
Is there any progress on the sum-of-three-cubes problem? I think 33 is the smallest number for which the answer is not known.
On Wed, May 30, 2018 at 5:29 PM, Mike Stay <metaweta@gmail.com> wrote:
:D https://cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.pdf
On Wed, May 30, 2018 at 2:54 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2018-05-29 09:59, Mike Stay wrote:
My favorite irrationality proof is one I heard from John Baez. Suppose the cube root of two were not irrational; then there would be two positive integers p, q such that p/q = ∛2. Multiplying both sides by q and cubing, we get p³ = 2q³ = q³ + q³, which has no solutions in the positive integers by Fermat's Last Theorem!
I just read this aloud to Rohan, who remarked "You can also use that for ∛9.", barely looking up from his video game. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.math.ucr.edu/~mike http://reperiendi.wordpress.com
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I remembered that somebody had asked, but I couldn't remember in what forum. I should have searched. And I don't think anybody mentioned, at the time, that the corresponding problem for 2 cubes was completely solved. (I believe that sums of 3 squares have been characterized completely, am I right?) On Wed, May 30, 2018 at 8:02 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Allan,
I just asked essentially the identical question to math-fun a few weeks ago, under a subject line something like "Unsolved number theory problem".
There were some interesting responses.
—Dan
-----Original Message-----
From: Allan Wechsler <acwacw@gmail.com> Sent: May 30, 2018 3:04 PM To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] "I don't find this proof beautiful" quote?
Is there any progress on the sum-of-three-cubes problem? I think 33 is the smallest number for which the answer is not known.
On Wed, May 30, 2018 at 5:29 PM, Mike Stay <metaweta@gmail.com> wrote:
:D https://cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.pdf
On Wed, May 30, 2018 at 2:54 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2018-05-29 09:59, Mike Stay wrote:
My favorite irrationality proof is one I heard from John Baez. Suppose the cube root of two were not irrational; then there would be two positive integers p, q such that p/q = ∛2. Multiplying both sides by q and cubing, we get p³ = 2q³ = q³ + q³, which has no solutions in the positive integers by Fermat's Last Theorem!
I just read this aloud to Rohan, who remarked "You can also use that for ∛9.", barely looking up from his video game. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.math.ucr.edu/~mike http://reperiendi.wordpress.com
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https://www.ams.org/journals/proc/1957-008-02/S0002-9939-1957-0085275-8/S000... A positive integer is the sum of three squares unless it is of the form 4^a (8n+7). -- Gene On Wednesday, May 30, 2018, 5:40:02 PM PDT, Allan Wechsler <acwacw@gmail.com> wrote: I remembered that somebody had asked, but I couldn't remember in what forum. I should have searched. And I don't think anybody mentioned, at the time, that the corresponding problem for 2 cubes was completely solved. (I believe that sums of 3 squares have been characterized completely, am I right?)
There’s a fun resemblance between the sums of squares game and the sums of palindromes game. In both cases, the number of special numbers up to N is about sqrt(N), arbitrarily large numbers can’t be written as the sum of three special numbers, and every sufficiently large number can be written as the sum of four special numbers. Here’s a somewhat forced way to continue the story, but maybe it leads somewhere. Given a modulus m such that Z/mZ has three cube roots of 1, say that an (m-1)-digit integer is a “3-drome” if the ith digit equals the jth digit whenever i^3 and j^3 are congruent mod m. The number of 3-dromes up to N is about N^(1/3), as is the number of positive perfect cubes up to N; are the additive number theory properties of the set of all 3-dromes analogous to the additive number theory properties of the set of all positive perfect cubes? My guess is “no”, as I expect that there are big gaps in the set of 3-dromes, corresponding to long runs of moduli m for which 1 does not have 3 cube roots. But maybe someone can see a way to fix my definition to make the analogy work. Jim Propp On Wednesday, May 30, 2018, Eugene Salamin via math-fun < math-fun@mailman.xmission.com> wrote:
https://www.ams.org/journals/proc/1957-008-02/S0002-9939- 1957-0085275-8/S0002-9939-1957-0085275-8.pdf
A positive integer is the sum of three squares unless it is of the form 4^a (8n+7).
-- Gene
On Wednesday, May 30, 2018, 5:40:02 PM PDT, Allan Wechsler < acwacw@gmail.com> wrote:
I remembered that somebody had asked, but I couldn't remember in what forum. I should have searched. And I don't think anybody mentioned, at the time, that the corresponding problem for 2 cubes was completely solved. (I believe that sums of 3 squares have been characterized completely, am I right?)
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I thought that sometime in the last year or so somebody posted a link to a paper showing that in bases larger than about 5, every number *is* the sum of three palindromes. On Wed, May 30, 2018 at 10:26 PM, James Propp <jamespropp@gmail.com> wrote:
There’s a fun resemblance between the sums of squares game and the sums of palindromes game. In both cases, the number of special numbers up to N is about sqrt(N), arbitrarily large numbers can’t be written as the sum of three special numbers, and every sufficiently large number can be written as the sum of four special numbers.
Here’s a somewhat forced way to continue the story, but maybe it leads somewhere. Given a modulus m such that Z/mZ has three cube roots of 1, say that an (m-1)-digit integer is a “3-drome” if the ith digit equals the jth digit whenever i^3 and j^3 are congruent mod m. The number of 3-dromes up to N is about N^(1/3), as is the number of positive perfect cubes up to N; are the additive number theory properties of the set of all 3-dromes analogous to the additive number theory properties of the set of all positive perfect cubes?
My guess is “no”, as I expect that there are big gaps in the set of 3-dromes, corresponding to long runs of moduli m for which 1 does not have 3 cube roots. But maybe someone can see a way to fix my definition to make the analogy work.
Jim Propp
On Wednesday, May 30, 2018, Eugene Salamin via math-fun < math-fun@mailman.xmission.com> wrote:
https://www.ams.org/journals/proc/1957-008-02/S0002-9939- 1957-0085275-8/S0002-9939-1957-0085275-8.pdf
A positive integer is the sum of three squares unless it is of the form 4^a (8n+7).
-- Gene
On Wednesday, May 30, 2018, 5:40:02 PM PDT, Allan Wechsler < acwacw@gmail.com> wrote:
I remembered that somebody had asked, but I couldn't remember in what forum. I should have searched. And I don't think anybody mentioned, at the time, that the corresponding problem for 2 cubes was completely solved. (I believe that sums of 3 squares have been characterized completely, am I right?)
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Oops! Allan is right. I had it wrong. My attempt at an analogy fails. Jim Propp On Wednesday, May 30, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
I thought that sometime in the last year or so somebody posted a link to a paper showing that in bases larger than about 5, every number *is* the sum of three palindromes.
On Wed, May 30, 2018 at 10:26 PM, James Propp <jamespropp@gmail.com> wrote:
There’s a fun resemblance between the sums of squares game and the sums of palindromes game. In both cases, the number of special numbers up to N is about sqrt(N), arbitrarily large numbers can’t be written as the sum of three special numbers, and every sufficiently large number can be written as the sum of four special numbers.
Here’s a somewhat forced way to continue the story, but maybe it leads somewhere. Given a modulus m such that Z/mZ has three cube roots of 1, say that an (m-1)-digit integer is a “3-drome” if the ith digit equals the jth digit whenever i^3 and j^3 are congruent mod m. The number of 3-dromes up to N is about N^(1/3), as is the number of positive perfect cubes up to N; are the additive number theory properties of the set of all 3-dromes analogous to the additive number theory properties of the set of all positive perfect cubes?
My guess is “no”, as I expect that there are big gaps in the set of 3-dromes, corresponding to long runs of moduli m for which 1 does not have 3 cube roots. But maybe someone can see a way to fix my definition to make the analogy work.
Jim Propp
On Wednesday, May 30, 2018, Eugene Salamin via math-fun < math-fun@mailman.xmission.com> wrote:
https://www.ams.org/journals/proc/1957-008-02/S0002-9939- 1957-0085275-8/S0002-9939-1957-0085275-8.pdf
A positive integer is the sum of three squares unless it is of the form 4^a (8n+7).
-- Gene
On Wednesday, May 30, 2018, 5:40:02 PM PDT, Allan Wechsler < acwacw@gmail.com> wrote:
I remembered that somebody had asked, but I couldn't remember in what forum. I should have searched. And I don't think anybody mentioned, at the time, that the corresponding problem for 2 cubes was completely solved. (I believe that sums of 3 squares have been characterized completely, am I right?)
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Replace “has three cube roots” by “has three or more cube roots” in what I wrote. Jim Propp On Wednesday, May 30, 2018, James Propp <jamespropp@gmail.com> wrote:
There’s a fun resemblance between the sums of squares game and the sums of palindromes game. In both cases, the number of special numbers up to N is about sqrt(N), arbitrarily large numbers can’t be written as the sum of three special numbers, and every sufficiently large number can be written as the sum of four special numbers.
Here’s a somewhat forced way to continue the story, but maybe it leads somewhere. Given a modulus m such that Z/mZ has three cube roots of 1, say that an (m-1)-digit integer is a “3-drome” if the ith digit equals the jth digit whenever i^3 and j^3 are congruent mod m. The number of 3-dromes up to N is about N^(1/3), as is the number of positive perfect cubes up to N; are the additive number theory properties of the set of all 3-dromes analogous to the additive number theory properties of the set of all positive perfect cubes?
My guess is “no”, as I expect that there are big gaps in the set of 3-dromes, corresponding to long runs of moduli m for which 1 does not have 3 cube roots. But maybe someone can see a way to fix my definition to make the analogy work.
Jim Propp
On Wednesday, May 30, 2018, Eugene Salamin via math-fun < math-fun@mailman.xmission.com> wrote:
https://www.ams.org/journals/proc/1957-008-02/S0002-9939-19 57-0085275-8/S0002-9939-1957-0085275-8.pdf
A positive integer is the sum of three squares unless it is of the form 4^a (8n+7).
-- Gene
On Wednesday, May 30, 2018, 5:40:02 PM PDT, Allan Wechsler < acwacw@gmail.com> wrote:
I remembered that somebody had asked, but I couldn't remember in what forum. I should have searched. And I don't think anybody mentioned, at the time, that the corresponding problem for 2 cubes was completely solved. (I believe that sums of 3 squares have been characterized completely, am I right?)
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participants (4)
-
Allan Wechsler -
Dan Asimov -
Eugene Salamin -
James Propp