[math-fun] ridiculously simple derivation (whose?) of Pythagorean triples
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)
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) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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)
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)
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-- 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
The same thing also comes out from |z| = |conj(z)| => 1^2 = |K + iL|^2 / |K - iL|^2 = |(K + iL) / (K - iL)|^2 = |(K + iL)^2 / (K^2 + L^2)|^2 = |(K^2 - L^2) + i2KL|^2 / (K^2 + L^2)^2 --Dan On 2013-04-26, at 1:04 AM, Bill Gosper 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)
P.S. Out of curiosity, I just plotted the surface z = (x^2 - y^2) / (x^2 + y^2) which is of course just z = cos(2 theta), and it gives a cute graph with a vertical interval as singular points at (x,y) = (0,0). This is well-known to some people, but it wasn't to me. You get essentially the identical graph for z = 2xy / (x^2 + y^2) = sin(2 theta). --Dan On 2013-04-26, at 1:06 PM, Dan Asimov wrote:
The same thing also comes out from
|z| = |conj(z)| =>
1^2 = |K + iL|^2 / |K - iL|^2 = |(K + iL) / (K - iL)|^2
= |(K + iL)^2 / (K^2 + L^2)|^2
= |(K^2 - L^2) + i2KL|^2 / (K^2 + L^2)^2
--Dan
On 2013-04-26, at 1:04 AM, Bill Gosper 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)
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participants (6)
-
Bill Gosper -
Dan Asimov -
Fred lunnon -
Henry Baker -
Neil Sloane -
Warut Roonguthai