2019/006) if $a^{x-1} = bc, b^{y-1} = ca, c^{z-1} = ab$ show that $xy + yz + zx = xyz$

From $a^{(x-1)} = bc$ we have $a^x= abc$ or $a=(abc)^\frac{1}{x}\cdots(1)$

similarly $b=(abc)^\frac{1}{y}\cdots(2)$
$c=(abc)^\frac{1}{z}\cdots(3)$

multiplying (1) (2) and (3) we get $abc = (abc)^{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$
or

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$

multiplying both sides by xyz we get the result 

2019/005) Prove If $gcd(a,b)=1$ then $gcd(a,b^2)=1$

We have $ax + by = 1$  as per bezout identity 

So$ b = 1 * b = b* (ax + by)$ 
$= bax + b^ 2 y$ 
So $by = baxy + b^2y^2$ 
Or $by + ax = baxy + ax + b^2 y^2$ 
Or $1 = ax(1+by) + b^2 y^2$ 

As we can put 1 as linear combination of $a$ and $b^2$ so $gcd(a,b^2) = 1$