Re: [math-fun] ridiculously simple derivation (whose?) of Pythagorean triples
Bumby! Excellent! Just exactly what I wanted. Thanks, Neil! http://www.math.rutgers.edu/~bumby/squares1.pdf At 06:26 AM 4/29/2013, Neil Sloane wrote:
Entry A000118 in the OEIS has many references. See, in particular, the Bumby reference.
On Mon, Apr 29, 2013 at 9:14 AM, Henry Baker <hbaker1@pipeline.com> wrote:
I'm still on the hunt for a 'ridiculously simple' way to express a positive integer as 4 squares.
Most presentations for the 4-square theorem focus on proving that it is possible, but don't provide very good algorithms for actually doing it.
At 09:47 AM 4/26/2013, Dan Asimov wrote:
How does that work with quaternions H or octonions O ?
You can't just switch one coefficient. (Or you can, but then you're just operating in a single complex subspace of H or O, so it's really the same thing.)
--Dan
On 2013-04-26, at 5:50 AM, Fred lunnon wrote:
Etc. for quaternions and octonions, where available ... WFL
On 4/26/13, Warut Roonguthai <warut822@gmail.com> wrote:
I don't know who discovered this, but I know that Pythagorean triples can be obtained from the fact that |z^2| = |z|^2. This is similar to the way you can prove that the product of numbers representable as sums of two squares is also representable as a sum of two squares by using the fact that |z1 * z2| = |z1| * |z2|.
On Fri, Apr 26, 2013 at 3:04 PM, Bill Gosper <billgosper@gmail.com> wrote:
In[570]:= (1+2*I)/(2+1*I) Out[570]= 4/5+(3 I)/5
In[571]:= (1+3*I)/(3+1*I) Out[571]= 3/5+(4 I)/5
In[572]:= (2+3*I)/(3+2*I) Out[572]= 12/13+(5 I)/13
In[573]:= (1+4*I)/(4+1*I) Out[573]= 8/17+(15 I)/17
In[586]:= Simplify/@ComplexExpand[(a+I*b)/(b+I*a)] Out[586]= (2 a b)/(a^2+b^2)-(I (a^2-b^2))/(a^2+b^2)
-- Dear Friends, I have now retired from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
Solving w^2 + x^2 + y^2 + z^2 = p , an odd prime, is rather simpler than Bumby manages to make it look. His final section mentions the Tonelli-Shanks algorithm for computing square root modulo p . So armed with this algorithm, compute a solution of a^2 + b^2 + c^2 = 0 (mod p) in time order log(p) roughly; then set Q = w + x i + y j + z k = GCD_L ( a i + b j + c k, p) , where GCD_L denotes left-hand quaternion GCD . Now ||Q|| = p , yielding a single solution. If all solutions are required, the simplest procedure is to continue by stepping around all p+1 points of the conic a^2 + b^2 + c^2 = 0 (mod p) in the projective plane over |F_p , starting from the solution found above, and converting each point via GCD with p as before. The resulting set {Q} comprises one representative from each unit-association class on the right: taking products with {1, i, j, k, -1, -i, -j, -k} yields the complete solution set in time order p log(p) , along with the cardinal 8(p+1) specified by Jacobi's theorem. Quaternion GCD is a natural extension of the Euclidean algorithm, slightly complicated by non-commutativity, and potential non-termination if both norms are even. Further details and program code are available. Fred Lunnon On 4/29/13, Henry Baker <hbaker1@pipeline.com> wrote:
Bumby! Excellent! Just exactly what I wanted. Thanks, Neil!
http://www.math.rutgers.edu/~bumby/squares1.pdf
At 06:26 AM 4/29/2013, Neil Sloane wrote:
Entry A000118 in the OEIS has many references. See, in particular, the Bumby reference.
On Mon, Apr 29, 2013 at 9:14 AM, Henry Baker <hbaker1@pipeline.com> wrote:
I'm still on the hunt for a 'ridiculously simple' way to express a positive integer as 4 squares.
Most presentations for the 4-square theorem focus on proving that it is possible, but don't provide very good algorithms for actually doing it.
At 09:47 AM 4/26/2013, Dan Asimov wrote:
How does that work with quaternions H or octonions O ?
You can't just switch one coefficient. (Or you can, but then you're just operating in a single complex subspace of H or O, so it's really the same thing.)
--Dan
On 2013-04-26, at 5:50 AM, Fred lunnon wrote:
Etc. for quaternions and octonions, where available ... WFL
On 4/26/13, Warut Roonguthai <warut822@gmail.com> wrote:
I don't know who discovered this, but I know that Pythagorean triples can be obtained from the fact that |z^2| = |z|^2. This is similar to the way you can prove that the product of numbers representable as sums of two squares is also representable as a sum of two squares by using the fact that |z1 * z2| = |z1| * |z2|.
On Fri, Apr 26, 2013 at 3:04 PM, Bill Gosper <billgosper@gmail.com> wrote:
> In[570]:= (1+2*I)/(2+1*I) > Out[570]= 4/5+(3 I)/5 > > In[571]:= (1+3*I)/(3+1*I) > Out[571]= 3/5+(4 I)/5 > > In[572]:= (2+3*I)/(3+2*I) > Out[572]= 12/13+(5 I)/13 > > In[573]:= (1+4*I)/(4+1*I) > Out[573]= 8/17+(15 I)/17 > > In[586]:= Simplify/@ComplexExpand[(a+I*b)/(b+I*a)] > Out[586]= (2 a b)/(a^2+b^2)-(I (a^2-b^2))/(a^2+b^2)
-- Dear Friends, I have now retired from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
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