This week, we will relax a little bit with a quite easy problem AQQ273:

Good luck and try to be among the first ones!
1/2.
Indeed: a/a = 2a, implies a is not 0 (we can't divide by 0), leaving: a = 2a^2, or 1=2a.
1/2.
Indeed: a/a = 2a, implies a is not 0 (we can't divide by 0), leaving: a = 2a^2, or 1=2a.
Under modular arithmetic (residue modulo) there is a unique solution to 1 = 2x (mod y) -- Gauss' notation
that is: x = (y+1)/2
So, since there is an infinite number of odd numbers, there is an infinite number of solutions, one for each (mod y), y being a odd integer.
y=3, x = (3+1)/2 = 2. Indeed 2x = 4 = 1 (mod 3) since 4 residue modulo 3 get a rest of 1.
y=5, x = (5+1)/2 = 3. Indeed 2x = 6 = 1 (mod 5),
etc. They all satisfy 2x = 1 (mod y)
Bingo vanderghast , you get it, congratulations (too easy for you...)!
Wow, I learn something... I was not aware about the modular arithmetic method...
Additionnal note: when y becomes larger, y+1 is approximatively equals to y and x/y tends toward the "continuous" solution, that is, 1/2. A little bit like quantum mech. (modular) tends toward classical mechanic.
I am really impressed (i.e. Schrondinger is part of our quizzes now..)