2017/014) if $x,y,z$ are in HP then show that $\log(x+z) + \log(x+z-2y) + 2 \log(x-z)$(IIT-1978)

let us eliminate y from LHS as RHS does not contain y
We have $x,y,z$ are in HP
hence $\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$
or $y(z+x) = 2xz$
or $y=\frac{2xz}{x+z}$
hence $x+z-2y = x+ z - \frac{4xz}{x+z}$
or $(x+z-2y)(x+z) = (x+z)^2 - 4xz = x^2+z^2+ 2xz - 4xz$
$ = x^2+z^2 - 2xz = (x-z)^2$
taking log on both sides we get $\log(x+z) + \log(x+z-2y) + 2 \log(x-z)$

2017/013) Prove that $\prod_{n=1}^{m-1} \sin \frac{n \pi}{m} = \frac{m}{2^{m-1}}$

we know $\sin \, nt = \frac{1}{2i}(e^{int} - e^{-int})=\frac{1}{2i}e^{-int}(e^{2int}-1) $
lettiing $t=\frac{\pi}{m}$ we have taking the product from n = to m-1 we get
$\prod_{n=1}^{m-1} sin \frac{n \pi}{m}= (\frac{1}{2i})^{m-1} e^{-i\sum_{n=1}^{m-1}\frac{n\pi}{m}}\prod_{n=1}^{m-1} (e^{2\frac{in\pi}{m}}-1) \cdots(1) $

Now $ e^{-i\sum_{n=1}^{m-1}\frac{n\pi}{m}} = e^\frac{-(m-1)(m)\pi\,i }{2m}= e^\frac{-(m-1)\pi\,i }{2}=  (e^\frac{-\pi\,i}{2})^{(m-1)}= (-i)^{m-1}\cdots(2)$

from (1) and (2)
$\prod_{n=1}^{m-1} sin \frac{n \pi}{m}= (\frac{1}{2i})^{m-1} (-i)^{m-1}= (\frac{-1}{2})^{m-1} \prod_{n=1}^{m-1} (e^{2\frac{in\pi}{m}}-1)$
$= (\frac{1}{2})^{m-1} \prod_{n=1}^{m-1} (1- e^{2\frac{in\pi}{m}}) = (\frac{1}{2})^{m-1} \prod_{n=1}^{m-1} (1- w^n) \dots (3)$ where w is $m^{th}$ root of 1

now because w is $n^{th}$ root of 1 we have

$x^n-1 = \prod_{n=0}^{m-1} (x- w^n) = (x-1)  \prod_{n=1}^{m-1} (x- w^n)\cdots(4)$
further $x^n-1 = (x-1) \sum_{n=0}^{m-1} x^n\cdots(5)$
so we have from (4) and (5)
$\prod_{n=1}^{m-1} (x- w^n) = \sum_{n=0}^{m-1} x^n$
putting x = 1 we get
$\prod_{n=1}^{m-1} (1- w^n) = \sum_{n=0}^{m-1} 1^n= m $

from (3) and above we get
$\prod_{n=1}^{m-1} \sin \frac{n \pi}{m} = \frac{m}{2^{m-1}}$

2017/012) Can $5^{64}-3^{64}$ be expressed as sum of 2 squares

we have $5^{64}-3^{64}= (5^{32} + 3^{32})(5^{32} - 3^{32})$
$= (5^{32} + 3^{32})(5^{16} + 3^{16})(5^{16} - 3^{16})$
$= (5^{32} + 3^{32})(5^{16} + 3^{16})(5^{8} + 3^{8})(5^{4} + 3^{4})(5^{4} - 3^{4})$
$= (5^{32} + 3^{32})(5^{16} + 3^{16})(5^{8} + 3^{8})(5^{4} + 3^{4})(5^{2} + 3^{2})(5^{2} - 3^{2})$
$= (5^{32} + 3^{32})(5^{16} + 3^{16})(5^{8} + 3^{8})(5^{4} + 3^{4})(5^{2} + 3^{2}) * 16$
the above is product of sum of squares( knowing that $16= 4^2 +0^2$ the above can be expressed as sum of squares