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Wednesday, June 26, 2019

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 

Monday, June 10, 2019

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