RMO 101 problems in algebra

 Q1.

Solution:

It gets 2 solutions 2/3 and 3/2.
=> (2/3)^x = 2/3,3/2 => x = -1,+1
Answer.

Q2.

Answer.

Q3.

So this is an A.P.
And X_n is an H.P.
After this:

Q4.

DS means diagonal sum here.

Key Insight: It's difficult to solve a degree 4 equation in 'x'. But it's easier to solve a degree 2 equation in 'a'.

Make a quadratic in 'a':

You will get 2 different quadratic equations:

Finally combine both solutions to get a >= 3/4

Key insight:
This is essentially 2 quadratic equations hidden in one.
We have to construct 2 separate equations.
One with x >=0 => |x| = x.
other with x <= 0. |x| = -x.

Now for them to have 3 distinct real solutions, there are 3 cases:
Case 1:
First equation has D = 0, second has D > 0
Case 2:
First equation has D > 0, second has D = 0
Case 3:
Both have D > 0 but one of the equations has only one valid root, i.e. in x >= 0 case, one root turns out to be negative.

Eventually you will get a = 1,3 as solutions.

First do some hit and trial to get the idea that x = -1 and y = -1 are solutions here.
So you can extract those factors out, i.e. (x+1),(y+1) and simplify.
Factorize it and solve:

One quick solution is (0,0,0).
Another is (-1,1,0).
Another is (-1,0,1) this suggests that x,y,z could be in A.P.
Also right hand side is built by differences. So A.P. could be a gateway here.

LHS is quadratic while RHS is cubic. So not simple.
Assuming A.P. reduces 3 variable equation to 2 variable equation.

If you assume the A.P. to be y-d,y,y+d you will eventually get a quadratic in y and d.
Once you put D = p.s. for y to be integer, you will get d = 6.n^2 + 1 where n is int. That will give you infinite values of x,y,z.