Dear Prof. Knuth, Toward the end of your most recent Christmas lecture, you presented the following conjecture, sent to you by Bill Gosper, which you showed to be true to at least 1000 decimal digits: $2^{\frac{2}{5}}\sqrt{5}\frac{{\Gamma \left(\frac{1}{5}\right)}^4}{{\Gamma \left(\frac{1}{10}\right)}^2{\Gamma \left(\frac{3}{10}\right)}^2}=\phi =\frac{1+\sqrt{5}}{2}$ Here's a proof which makes use of the following three prior results:
(a)	Euler's reflection formula for the gamma function (for non-integer z): $\Gamma \left(z\right)\Gamma (1-z)=\frac{\pi }{\sin (\pi z)}$
(b)	The Legendre duplication formula for the gamma function: $\Gamma \left(z\right)\Gamma (z+\frac{1}{2})=2^{1-2z}\sqrt{\pi }\Gamma (2z)$
(c)	The following expression which can be derived from the multiple-angle formula for sin(5x) that uses the Chebyshev polynomial of the first kind (see http://mathworld.wolfram.com /TrigonometryAnglesPi5.html which uses equation 18 from http://mathworld.wolfram.com /Multiple-AngleFormulas.html): $\sin \left(\frac{\pi }{5}\right)=\frac{1}{4}\sqrt{10-2\sqrt{5}}$

Proof:

(1)	Applying (a) (for z=1/5) followed by (c), we get $\Gamma \left(\frac{1}{5}\right)\Gamma \left(\frac{4}{5}\right)=\frac{\pi }{\sin \left(\frac{\pi }{5}\right)}=\frac{4\pi }{\sqrt{10-2\sqrt{5}}}$
(2)	Applying (b) for z=3/10 and z=1/10, we get $\Gamma \left(\frac{3}{10}\right)\Gamma \left(\frac{8}{10}\right)=2^{\frac{2}{5}}\sqrt{\pi }\Gamma \left(\frac{6}{10}\right)$ and $\Gamma \left(\frac{1}{10}\right)\Gamma \left(\frac{6}{10}\right)=2^{\frac{4}{5}}\sqrt{\pi }\Gamma \left(\frac{2}{10}\right)$ Therefore: $\Gamma \left(\frac{8}{10}\right)=\Gamma \left(\frac{4}{5}\right)=2^{\frac{6}{5}}\pi \frac{\Gamma \left(\frac{1}{5}\right)}{\Gamma \left(\frac{1}{10}\right)\Gamma \left(\frac{3}{10}\right)}$
(3)	Substituting the result from step 2 into step 1, we get $2^{\frac{6}{5}}\pi \frac{{\Gamma \left(\frac{1}{5}\right)}^2}{\Gamma \left(\frac{1}{10}\right)\Gamma \left(\frac{3}{10}\right)}=\frac{4\pi }{\sqrt{10-2\sqrt{5}}}$
(4)	Dividing both sides by $\pi$ and squaring both sides, we get $2^{\frac{12}{5}}\frac{{\Gamma \left(\frac{1}{5}\right)}^4}{{\Gamma \left(\frac{1}{10}\right)}^2{\Gamma \left(\frac{3}{10}\right)}^2}=\frac{16}{10-2\sqrt{5}}=\frac{8}{5-\sqrt{5}}$
(5)	After multiplying both sides by $\sqrt{5}$ and simplifying, we're done: $2^{\frac{2}{5}}\sqrt{5}\frac{{\Gamma \left(\frac{1}{5}\right)}^4}{{\Gamma \left(\frac{1}{10}\right)}^2{\Gamma \left(\frac{3}{10}\right)}^2}=\frac{2\sqrt{5}}{5-\sqrt{5}}\left(\frac{5+\sqrt{5}}{5+\sqrt{5}}\right)=\frac{10\sqrt{5}+10}{20}=\frac{\sqrt{5}+1}{2}$

Best regards,
Pius Fischer
March 4, 2018