arbitrary factorization as a difference of cubes?
Kieren
Hello!
It is well known that every number can be factored as a difference of
squares, as seen by the identity
ab = ((a+b)/2)^2 - ((a-b)/2)^2.
Are there any equivalent identities which use a difference of cubes?
Many thanks,
Kieren.
Wolfgang Rave
On Sep 15, 10:13?pm, Kieren <kie...@alumni.rice.edu> wrote:
> Hello!
>
> It is well known that every number can be factored as a difference of
> squares, as seen by the identity
>
> ? ? ab = ((a+b)/2)^2 - ((a-b)/2)^2.
>
> Are there any equivalent identities which use a difference of cubes?
>
> Many thanks,
> Kieren.
hi Kieren,
the above identity is based on a^2 - b^2 = (a+b)*(a-b). The analogon
for cubes is given by
a^3 pm b^3 = (a pm b)*(a^2 mp ab + b^2), where pm means plus/minus and
mp vice versa.
hope this helps, Wolfgang.
Kieren
Hi Wolfgang,
Thanks for the post. I *do* know that analog for cubes -- allow me to
clarify my question?
Let's write
ab = F_1(a,b)^2 - F_2(a,b)^2
where F_1(a,b) = (a+b)/2 and F_2(a,b) = (a-b)/2. Then I want an
analogous result for cubes, i.e.
ab = F_1(a,b)^3 +- F_2(a,b)^3 +- F_3(a,b)^3 +- ...
where the F_k(a,b) are functions/polynomials in a and b. Since (e.g.)
it is known that every rational number is the sum of three positive
rational cubes [q.v. Hardy & Wright, Dickson's History, etc.], I
figure it can be done in three terms/functions (although maybe not in
*simple* functions of a and b?).
I'm happy to hack away at trying to derive the F_k myself, using the
H&W/Dickson solution(s) as starting points -- but I thought I'd save
myself some wheel-reinvention if anyone out there already knew the
answer!
Thanks,
Kieren.
Dave Rusin
In article <gamfm0$1d1$1...@dhcp-94-62.math.ohio-state.edu>,
Kieren <kie...@alumni.rice.edu> wrote:
>It is well known that every number can be factored as a difference of
>squares, as seen by the identity
>
> ab = ((a+b)/2)^2 - ((a-b)/2)^2.
I'm not quite I understand your interpretation of this identity.
I might interpret it as follows: given an integer N we consider
two Diophantine equations to be solved: N = a b and N = x^2 - y^2.
(Or more generally we can play this game in any ring, not just in Z .)
>From each solution (x,y) to the latter equation, we obtain a solution
(a,b) to the former, namely a = x+y, b = x-y. We can ask whether
this mapping is one-to-one (it is) and whether it is onto (it's not,
although it is onto whenever 1/2 is in the ring, as your identity shows;
for the purposes of practical integer factorization that's good enough
since we are probably only interested in the case where N is odd).
>Are there any equivalent identities which use a difference of cubes?
Continuing the interpretation I gave above, we now consider the
solution sets to the equations N = a b and N = x^3 - y^3 . There
is again a mapping from the second to the first, giving
(a,b)=(x-y, x^2+xy+y^2). This time it's not quite one-to-one since
(x,y) and (-y,-x) give the same pair (a,b). But more significantly,
it's not even close to being onto: when a pair (a,b) is given and it
does correspond to a pair (x,y), then we note that 12 b - 3 a^2 =
(3(x+y))^2 would have to be a perfect square. Obviously this does
not happen for all (a,b) (unless N is very special). For example,
143 = 11 * 13 but there is no way to write 143 as a difference of
two cubes -- even if they are allowed to be rational or real --
such that 11 = x-y and 13 = x^2+xy+y^2 .
(Apart from some congruences mod 2 and 3, akin to the case with squares,
this is the only obstruction for cubes: every factorization N = a b
for which 12b - 3a^2 is a square arises from a presentation of N
as a difference of two cubes.)
dave
Kieren
Hello, all!
Thank you for your responses (both on- and off-list). Sorry I
apparently wasn't any clearer with my "clarification" post than I was
in my original? =)
In any case, I have (in the meantime) derived an identity which is an
example of precisely the kind of thing I was seeking:
ab = F_1(a,b)^3 +- F_2(a,b)^3 +- F_3(a,b)^3 +- ...
is satisfied by
F_1(a,b) = (27a^3 - 81a^2b^2 + 45ab^4 + b^6)/(6b(3a+b^2)^2)
F_2(a,b) = (6ab(3a-b^2))/(3a+b^2)^2
F_3(a,b) = -(9a^2 - 30ab^2 + b^4)/(6b(3a+b^2)
which is to say
ab = [(27a^3 - 81a^2b^2 + 45ab^4 + b^6)/(6b(3a+b^2)^2)]^3 +
[(6ab(3a-b^2))/(3a+b^2)^2]^3 - [(9a^2 - 30ab^2 + b^4)/(6b(3a+b^2)]^3.
Hopefully that example makes my original post clearer? If so, does
anyone know any *other* algebraic identities of this type?
Thanks,
Kieren.
Dave L. Renfro
> It is well known that every number can be factored as
> a difference of squares, as seen by the identity
>
> ab = ((a+b)/2)^2 - ((a-b)/2)^2.
>
> Are there any equivalent identities which use a difference
> of cubes?
For what it's worth, the uses for this identity I've seen
are based on the fact that it provides a way to write the
product ab in terms of additions, subtractions, division
by a simple constant (2), and squaring. Here are a couple of
places that I can think of where this it's used: (1) In numerical
calculations by hand, before electronic computing devices
were developed (see most any 19th century algebra book,
the sections where numerical computations are discussed).
(2) In showing that a product of Lebesgue measurable functions
is Lebesgue measurable (by first showing sums, differences,
constant multiples, and squares are measurable via the
Caratheodory definition).
The identity can be written as
4ab = (a+b)^2 - (a-b)^2
Yesterday I happened to come across a similar identity
for the product of three factors:
24abc = (a+b+c)^3 - (a+b-c)^3 - (a-b+c)^3 - (-a+b+c)^3
Although I would imagine this can be found in many
algebra texts from the 19th century, I came across
this identity in the following short note:
[no author given], "Uebungsaufgaben für Schüler"
[Exercises for Schools], Archiv der Mathematik und
Physik (1) 6 (1845), 105-106.
http://books.google.com/books?id=k50EAAAAQAAJ&pg=PA105
I don't have a computer algebra system available to
verify this, but I believe it can be verified without
much difficulty using the factor theorem one encounters
in (present day) college algebra and precalculus courses.
(A couple of minutes later ... Yes, this approach works.)
Consider the expression on the right hand side above:
** If a = 0, we get (a+b)^3 - (b-c)^3 - (-b+c)^3 - (b+c)^3 = 0.
** If b = 0, we get (a+c)^3 - (a-c)^3 - (a+c)^3 - (-a+c)^3 = 0.
** If c = 0, we get (a+b)^3 - (a+b)^3 - (a-b)^3 - (-a+b)^3 = 0.
Therefore, a - 0 is a factor of the right hand side, b - 0
is a factor of the right hand side, and c - 0 is a factor
of the right hand side. Hence, abc is a factor of the right
hand side. Since it is clear that the right hand side has
degree (at most) 3, it follows that the right hand side
is a constant times abc. We can find the value of this
constant by substituting a = b = c = 1, which gives:
(constant)(1)(1)(1) = 3^3 - 1^3 - 1^3 - 1^3
constant = 24
Thus, the right hand side equals 24abc, which verifies
the identity.
Dave L. Renfro
mjc
>
> I don't have a computer algebra system available to
> verify this, but I believe it can be verified without
> much difficulty using the factor theorem one encounters
> in (present day) college algebra and precalculus courses.
> (A couple of minutes later ... Yes, this approach works.)
>
> Consider the expression on the right hand side above:
>
> ** If a = 0, we get (a+b)^3 - (b-c)^3 - (-b+c)^3 - (b+c)^3 = 0.
>
> ** If b = 0, we get (a+c)^3 - (a-c)^3 - (a+c)^3 - (-a+c)^3 = 0.
>
> ** If c = 0, we get (a+b)^3 - (a+b)^3 - (a-b)^3 - (-a+b)^3 = 0.
>
> Therefore, a - 0 is a factor of the right hand side, b - 0
> is a factor of the right hand side, and c - 0 is a factor
> of the right hand side. Hence, abc is a factor of the right
> hand side. Since it is clear that the right hand side has
> degree (at most) 3, it follows that the right hand side
> is a constant times abc. We can find the value of this
> constant by substituting a = b = c = 1, which gives:
>
> (constant)(1)(1)(1) = 3^3 - 1^3 - 1^3 - 1^3
>
> constant = 24
>
> Thus, the right hand side equals 24abc, which verifies
> the identity.
>
> Dave L. Renfro
As a trivial aside, it can be written as
24abc = (a+b+c)^3 + (-a-b+c)^3 + (-a+b-c)^3 + (a-b-c)^3.
Robert Israel
> [no author given], "Uebungsaufgaben f�r Sch�ler"
> [Exercises for Schools], Archiv der Mathematik und
> Physik (1) 6 (1845), 105-106.
> http://books.google.com/books?id=k50EAAAAQAAJ&pg=PA105
>
> I don't have a computer algebra system available to
> verify this, but I believe it can be verified without
> much difficulty using the factor theorem one encounters
> in (present day) college algebra and precalculus courses.
> (A couple of minutes later ... Yes, this approach works.)
>
> Consider the expression on the right hand side above:
>
> ** If a = 0, we get (a+b)^3 - (b-c)^3 - (-b+c)^3 - (b+c)^3 = 0.
>
> ** If b = 0, we get (a+c)^3 - (a-c)^3 - (a+c)^3 - (-a+c)^3 = 0.
>
> ** If c = 0, we get (a+b)^3 - (a+b)^3 - (a-b)^3 - (-a+b)^3 = 0.
>
> Therefore, a - 0 is a factor of the right hand side, b - 0
> is a factor of the right hand side, and c - 0 is a factor
> of the right hand side. Hence, abc is a factor of the right
> hand side. Since it is clear that the right hand side has
> degree (at most) 3, it follows that the right hand side
> is a constant times abc. We can find the value of this
> constant by substituting a = b = c = 1, which gives:
>
> (constant)(1)(1)(1) = 3^3 - 1^3 - 1^3 - 1^3
>
> constant = 24
>
> Thus, the right hand side equals 24abc, which verifies
> the identity.
Along the same lines,
(a+b+c+d)^4 - (a+b+c-d)^4 - (a+b-c+d)^4 - (a-b+c+d)^4 - (-a+b+c+d)^4
+(a+b-c-d)^4 + (a-b+c-d)^4 + (a-b-c+d)^4 = 192 a b c d
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Dan Hoey
Robert Israel wrote:
> Along the same lines,
>
> (a+b+c+d)^4 - (a+b+c-d)^4 - (a+b-c+d)^4 - (a-b+c+d)^4 - (-a+b+c+d)^4
> +(a+b-c-d)^4 + (a-b+c-d)^4 + (a-b-c+d)^4 = 192 a b c d
Indeed, it appears that
(#P)! \product_(p \in P) 2p
=? Sum_(Q \subset P) (-1)^#Q ( Sum_(p \in P) PM(p,Q) )^#P,
where PM(p,Q) = -p if p \in Q, p otherwise.
There should be a name for this equation, but I don't know it.
Dan Hoey
haoyuep at aol.com