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sum of two consecutive cubes can be prime?
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Gerard Schildberger
Nov 9, 2012, 2:57:53 PM11/9/12
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Can the sum of two consecutive cubes be a prime?
"Cube" can be any integer that is an integer raised to the 3rd power,
or in other words, a perfect cube.
__________________________________________________ Gerard Schildberger
"Cube" can be any integer that is an integer raised to the 3rd power,
or in other words, a perfect cube.
__________________________________________________ Gerard Schildberger
Brian M. Scott
Nov 9, 2012, 4:37:04 PM11/9/12
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On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger
<gera...@rrt.net> wrote in
<news:2c0b2bf7-b22d-4833...@googlegroups.com>
in alt.math.recreational:
> Can the sum of two consecutive cubes be a prime?
No. Neither can the sum of two non-consecutive cubes.
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
Brian
<gera...@rrt.net> wrote in
<news:2c0b2bf7-b22d-4833...@googlegroups.com>
in alt.math.recreational:
> Can the sum of two consecutive cubes be a prime?
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
Brian
Gerard Schildberger
Nov 9, 2012, 6:27:10 PM11/9/12
to b.s...@csuohio.edu
On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger wrote:
> <news:2c0b2bf7-b22d-4833...@googlegroups.com>
> in alt.math.recreational:
>> Can the sum of two consecutive cubes be a prime?
> No. Neither can the sum of two non-consecutive cubes.
> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> Brian
The trivial case of A=1 and B=1. _____________________ Gerard S.
> <news:2c0b2bf7-b22d-4833...@googlegroups.com>
> in alt.math.recreational:
>> Can the sum of two consecutive cubes be a prime?
> No. Neither can the sum of two non-consecutive cubes.
> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> Brian
Gerard Schildberger
Nov 9, 2012, 6:37:43 PM11/9/12
to b.s...@csuohio.edu
On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger
> <news:2c0b2bf7-b22d-4833...@googlegroups.com>
> in alt.math.recreational:
>> Can the sum of two consecutive cubes be a prime?
> No. Neither can the sum of two non-consecutive cubes.
> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> Brian
A=2 B=-1 I assume there must be others? _________ Gerard S.
> in alt.math.recreational:
>> Can the sum of two consecutive cubes be a prime?
> No. Neither can the sum of two non-consecutive cubes.
> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> Brian
Brian M. Scott
Nov 9, 2012, 7:20:24 PM11/9/12
to
On Fri, 9 Nov 2012 15:27:10 -0800 (PST), Gerard Schildberger
<gera...@rrt.net> wrote in
<news:0a538eec-6fa1-4471...@googlegroups.com>
in alt.math.recreational:
> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger wrote:
>> <news:2c0b2bf7-b22d-4833...@googlegroups.com>
>> in alt.math.recreational:
>>> Can the sum of two consecutive cubes be a prime?
>> No. Neither can the sum of two non-consecutive cubes.
>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
<gera...@rrt.net> wrote in
<news:0a538eec-6fa1-4471...@googlegroups.com>
in alt.math.recreational:
> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger wrote:
>> <news:2c0b2bf7-b22d-4833...@googlegroups.com>
>> in alt.math.recreational:
>>> Can the sum of two consecutive cubes be a prime?
>> No. Neither can the sum of two non-consecutive cubes.
>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> The trivial case of A=1 and B=1. _____________________ Gerard S.
Obviously. I assumed that it did not require mention.
Brian M. Scott
Nov 9, 2012, 7:21:21 PM11/9/12
to
On Fri, 9 Nov 2012 15:37:43 -0800 (PST), Gerard Schildberger
<gera...@rrt.net> wrote in
<news:99b221ed-d8ff-4310...@googlegroups.com>
in alt.math.recreational:
> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger
>> <news:2c0b2bf7-b22d-4833...@googlegroups.com>
>> in alt.math.recreational:
>>> Can the sum of two consecutive cubes be a prime?
>> No. Neither can the sum of two non-consecutive cubes.
>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
of a and b, you should have said so; the default
assumption in this context is otherwise.
<gera...@rrt.net> wrote in
<news:99b221ed-d8ff-4310...@googlegroups.com>
in alt.math.recreational:
> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger
>> <news:2c0b2bf7-b22d-4833...@googlegroups.com>
>> in alt.math.recreational:
>>> Can the sum of two consecutive cubes be a prime?
>> No. Neither can the sum of two non-consecutive cubes.
>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> A=2 B=-1 I assume there must be others? _________ Gerard S.
If you meant to include negative integers as possible values
of a and b, you should have said so; the default
assumption in this context is otherwise.
William Elliot
Nov 9, 2012, 8:48:05 PM11/9/12
to
On Fri, 9 Nov 2012, Gerard Schildberger wrote:
> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>
> >> Can the sum of two consecutive cubes be a prime?
>
> > No. Neither can the sum of two non-consecutive cubes.
> > a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>
> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>
> >> Can the sum of two consecutive cubes be a prime?
>
> > No. Neither can the sum of two non-consecutive cubes.
> > a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>
> A=2 B=-1 I assume there must be others? _________ Gerard S.
>
Assume a,b in Z. When is
>
a^3 + b^3 = (a + b)(a^2 - ab + b^2) a prime?
Cases are:
a = b = +-1,
a = 1 - b; a^2 - a(1 - a) + (1 - a)^2 = 3a^2 - 3a + 1 is prime.
Gerard Schildberger
Nov 9, 2012, 9:32:28 PM11/9/12
to b.s...@csuohio.edu
On Friday, November 9, 2012 6:21:27 PM UTC-6, Brian M. Scott wrote:
> On Fri, 9 Nov 2012 15:37:43 -0800 (PST), Gerard Schildberger wrote
after my post, I assume you meant me. I mentioned two consecutive
cubes (by definition, integers), and I didn't exclude zero or negative
numbers. I didn't post the equation, so it wasn't me who excluded
negative integers, I only supplied an example of a case where a
negative integer disproves the equation which had no condition which
mentioned that a, b ≥ 0. I made no assumptions about the
bounds for the integers. ________________________ Gerard Schildberger
> On Fri, 9 Nov 2012 15:37:43 -0800 (PST), Gerard Schildberger wrote
> in alt.math.recreational:
>> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>>> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger wrote
>> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>>> in alt.math.recreational:
>>>> Can the sum of two consecutive cubes be a prime?
>>> No. Neither can the sum of two non-consecutive cubes.
>>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>> A=2 B=-1 I assume there must be others? _________ Gerard S.
> If you meant to include negative integers as possible values
> of a and b, you should have said so; the default
> assumption in this context is otherwise.
It's unclear to me who you meant by "you", but since it posted right
>>>> Can the sum of two consecutive cubes be a prime?
>>> No. Neither can the sum of two non-consecutive cubes.
>>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>> A=2 B=-1 I assume there must be others? _________ Gerard S.
> If you meant to include negative integers as possible values
> of a and b, you should have said so; the default
> assumption in this context is otherwise.
after my post, I assume you meant me. I mentioned two consecutive
cubes (by definition, integers), and I didn't exclude zero or negative
numbers. I didn't post the equation, so it wasn't me who excluded
negative integers, I only supplied an example of a case where a
negative integer disproves the equation which had no condition which
mentioned that a, b ≥ 0. I made no assumptions about the
bounds for the integers. ________________________ Gerard Schildberger
Brian M. Scott
Nov 9, 2012, 10:51:00 PM11/9/12
to
On Fri, 9 Nov 2012 18:32:28 -0800 (PST), Gerard Schildberger
<gera...@rrt.net> wrote in
<news:14c2b814-7abe-42f8...@googlegroups.com>
up near the top of this one:
>> On Fri, 9 Nov 2012 15:37:43 -0800 (PST), Gerard Schildberger wrote
>> in alt.math.recreational:
Moreover, the References header in my post shows that it's
threaded to yours. Try using a real newsreader instead of
that crippled interface provided by Google Groups.
> but since it posted right after my post, I assume you
> meant me. I mentioned two consecutive cubes (by
> definition, integers), and I didn't exclude zero or
> negative numbers.
Not explicitly, no. The point is that a question about
primes carries that as a contextual implication, so if that
implication is not wanted, you should say so.
Possibly you're a bit dim and failed to recognize not only
the context that you were implying with your question but
also the assumption that my answer obviously entailed; your
inability to recognize that my post was unambiguously a
response to you tends to support this possibility.
Possibly you just want to argue.
Possibly both; the two are hardly mutually exclusive.
Whatever your problem, I now regret my attempt to be helpful
and will leave you to your own devices.
<gera...@rrt.net> wrote in
<news:14c2b814-7abe-42f8...@googlegroups.com>
in alt.math.recreational:
> On Friday, November 9, 2012 6:21:27 PM UTC-6, Brian M. Scott wrote:
>> On Fri, 9 Nov 2012 15:37:43 -0800 (PST), Gerard Schildberger wrote
>> in alt.math.recreational:
>>> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>>>> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger wrote
>>>> in alt.math.recreational:
>>>>> Can the sum of two consecutive cubes be a prime?
>>>> No. Neither can the sum of two non-consecutive cubes.
>>>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>>> A=2 B=-1 I assume there must be others? _________ Gerard S.
>> If you meant to include negative integers as possible values
>> of a and b, you should have said so; the default
>> assumption in this context is otherwise.
> It's unclear to me who you meant by "you",
It shouldn't be: the first lines of my post are still quoted
> On Friday, November 9, 2012 6:21:27 PM UTC-6, Brian M. Scott wrote:
>> On Fri, 9 Nov 2012 15:37:43 -0800 (PST), Gerard Schildberger wrote
>> in alt.math.recreational:
>>> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>>>> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger wrote
>>>> in alt.math.recreational:
>>>>> Can the sum of two consecutive cubes be a prime?
>>>> No. Neither can the sum of two non-consecutive cubes.
>>>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>>> A=2 B=-1 I assume there must be others? _________ Gerard S.
>> If you meant to include negative integers as possible values
>> of a and b, you should have said so; the default
>> assumption in this context is otherwise.
> It's unclear to me who you meant by "you",
up near the top of this one:
>> On Fri, 9 Nov 2012 15:37:43 -0800 (PST), Gerard Schildberger wrote
>> in alt.math.recreational:
threaded to yours. Try using a real newsreader instead of
that crippled interface provided by Google Groups.
> but since it posted right after my post, I assume you
> meant me. I mentioned two consecutive cubes (by
> definition, integers), and I didn't exclude zero or
> negative numbers.
primes carries that as a contextual implication, so if that
implication is not wanted, you should say so.
Possibly you're a bit dim and failed to recognize not only
the context that you were implying with your question but
also the assumption that my answer obviously entailed; your
inability to recognize that my post was unambiguously a
response to you tends to support this possibility.
Possibly you just want to argue.
Possibly both; the two are hardly mutually exclusive.
Whatever your problem, I now regret my attempt to be helpful
and will leave you to your own devices.
Frederick Williams
Nov 10, 2012, 4:46:49 AM11/10/12
to
"Brian M. Scott" wrote:
>
> On Fri, 9 Nov 2012 15:27:10 -0800 (PST), Gerard Schildberger
> <gera...@rrt.net> wrote in
> <news:0a538eec-6fa1-4471...@googlegroups.com>
> in alt.math.recreational:
>
> > On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>
> >> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger wrote:
> >> <news:2c0b2bf7-b22d-4833...@googlegroups.com>
> >> in alt.math.recreational:
>
> >>> Can the sum of two consecutive cubes be a prime?
>
> >> No. Neither can the sum of two non-consecutive cubes.
> >> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>
> > The trivial case of A=1 and B=1. _____________________ Gerard S.
>
> Obviously. I assumed that it did not require mention.
Doesn't "two consecutive" make mention of it redundant?
>
> On Fri, 9 Nov 2012 15:27:10 -0800 (PST), Gerard Schildberger
> <gera...@rrt.net> wrote in
> <news:0a538eec-6fa1-4471...@googlegroups.com>
> in alt.math.recreational:
>
> > On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>
> >> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger wrote:
> >> <news:2c0b2bf7-b22d-4833...@googlegroups.com>
> >> in alt.math.recreational:
>
> >>> Can the sum of two consecutive cubes be a prime?
>
> >> No. Neither can the sum of two non-consecutive cubes.
> >> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>
> > The trivial case of A=1 and B=1. _____________________ Gerard S.
>
> Obviously. I assumed that it did not require mention.
--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting
Frederick Williams
Nov 10, 2012, 4:48:32 AM11/10/12
to
Frederick Williams
Nov 10, 2012, 4:53:47 AM11/10/12
to
the 3rd power...' Surely nowadays "integer" includes negative numbers?
I accept that "prime" implies positive, so dropping "consecutive", 8 +
-1 seems to be a solution.
Gerard Schildberger
Nov 10, 2012, 12:58:10 PM11/10/12
to
On Saturday, November 10, 2012 3:48:35 AM UTC-6, Frederick Williams wrote:
> Gerard Schildberger wrote:
>> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>>> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger wrote
> Gerard Schildberger wrote:
>> On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott wrote:
>>> in alt.math.recreational:
>>>> Can the sum of two consecutive cubes be a prime?
>>> No. Neither can the sum of two non-consecutive cubes.
>>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>>> Brian
>> A=2 B=-1 I assume there must be others? _________ Gerard S.
> What does "consecutive" mean to you?
I was responding the the equation which uses the phrase:
>>>> Can the sum of two consecutive cubes be a prime?
>>> No. Neither can the sum of two non-consecutive cubes.
>>> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
>>> Brian
>> A=2 B=-1 I assume there must be others? _________ Gerard S.
> What does "consecutive" mean to you?
non-consecutive cubes
and gave two counterexamples. __________________ Gerard S.
Frederick Williams
Nov 10, 2012, 1:28:46 PM11/10/12
to
Peter Webb
Nov 13, 2012, 10:10:13 AM11/13/12
to
Frederick Williams wrote:
> "Brian M. Scott" wrote:
> >
> > On Fri, 9 Nov 2012 15:27:10 -0800 (PST), Gerard Schildberger
> > <gera...@rrt.net> wrote in
> > <news:0a538eec-6fa1-4471...@googlegroups.com>
> > in alt.math.recreational:
> >
> > > On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott
> > > wrote:
> >
> > >> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger
> > wrote: >>
> > <news:2c0b2bf7-b22d-4833...@googlegroups.com> >> in
> > alt.math.recreational:
> >
> > >>> Can the sum of two consecutive cubes be a prime?
> >
> > >> No. Neither can the sum of two non-consecutive cubes.
> > >> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> >
> > > The trivial case of A=1 and B=1. _____________________ Gerard
> > > S.
> >
> > Obviously. I assumed that it did not require mention.
>
> Doesn't "two consecutive" make mention of it redundant?
No
> "Brian M. Scott" wrote:
> >
> > On Fri, 9 Nov 2012 15:27:10 -0800 (PST), Gerard Schildberger
> > <gera...@rrt.net> wrote in
> > <news:0a538eec-6fa1-4471...@googlegroups.com>
> > in alt.math.recreational:
> >
> > > On Friday, November 9, 2012 3:37:11 PM UTC-6, Brian M. Scott
> > > wrote:
> >
> > >> On Fri, 9 Nov 2012 11:57:53 -0800 (PST), Gerard Schildberger
> > wrote: >>
> > <news:2c0b2bf7-b22d-4833...@googlegroups.com> >> in
> > alt.math.recreational:
> >
> > >>> Can the sum of two consecutive cubes be a prime?
> >
> > >> No. Neither can the sum of two non-consecutive cubes.
> > >> a^3 + b^3 = (a + b)(a^2 - ab + b^2)
> >
> > > The trivial case of A=1 and B=1. _____________________ Gerard
> > > S.
> >
> > Obviously. I assumed that it did not require mention.
>
> Doesn't "two consecutive" make mention of it redundant?
Sum of consecutive primes could be 1 + 8 + 27 + 64
Sum of two primes could be 1 + 64
2321...@gmail.com
Jan 8, 2013, 8:09:12 AM1/8/13
to
This can be factored to (2n+1)(n^2+n+1)
Example: if n=3, 3^3+4^3=27+64=91 or (7)(19)
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