Q2020/14) Solve in natural numbers $\frac{1}{a} + \frac{1}{b} = \frac{3}{2018}$

we get
$2018b + 2018 a = 3ab$
or $3ab - 2018 a - 2018b = 0$
As the coefficient of a,b are same so we get product of 2 monomials one of a one of b and coefficients being same.
Multiplying by 3 we get
$9ab - 3 * 2018 a - 3 * 2018b = 0$
or $9ab - 3 * 2018 a - 3 * 2018b = 2018 ^2 $
or $(3a-2018)(3b-2018) = 2018^2= 2^2 * 1009^2$
LHS each term is 1 mod 3 so we should make the RHS so product of 2 numbers each of which is 1 mod 3.


if(a,b) is a solution then (b,a) is also a solution so let us assume that $a>=b$

so we get following sets

$3a-2018= 2018^2, 3b-2018=1$  giving $(a= 1358114,b= 673)$
$3a-2018= 1009^2 , 3b=2018=4$ giving $(a= 340033, b= 674)$
$3a-2018= 1009, 3b-2018=1009 * 4$ giving $(a= 1009,b= 2018)$

and permutation of the same.

2020/13) Solve $2\log_2(x+15) - \log_2x = 6$

We have
$2\log_2(x+15) - \log_2x = \log_2(x+15)^2 - \log_2(x) = \log_2\frac{(x+15)^2}{x} = 6$
Or $\frac{(x+15)^2}{x} = 2^6= 64$
Or $(x+15)^2 = 64x$
Or $x^2+30x+225-64x=0$
Or $x^2-34x+ 225=0$
Or $(x-25)(x-9)= 0$ So x = 25 or 9 

2020/012) Solve in real $x^4+y^4+z^4 + 1 = xyz$

We have LHS is positive so RHS is also positive.

So all of x,y,z are positive or one is positive and rest 2 are negative

So let us solve for positive x,y,z

Using AM GM inequality we have  $\frac{x^4+y^4+z^4+1}{4} >=\sqrt[4]{x^4y^4z^4}$

Or $x^4+y^4+z^4 +1 >= 4xyz$

And they are equal if $x=y=z$

So we get $3x^4-4x^3+1 = 0$

Or  we have trying $x=1$ and $x = 3$ x=1 is solution

giving one set of solution = $(1,1,1)$ and 2 negative  and one positive gives 3 more $(1,-1,-1), (-1,-1,1), \,and\, (-1,1,-1)$