PLSSSSS HELPPPPP function question (((PLSSSS ITS URGENTTT))))

\(f(x) = f(f(x))\)

\(x^2 - 2x = (x^2-2x)^2 - 2(x^2-2x)\)

Let \(x^2 - 2x = y.\)

\(y = y^2 - 2y\)

\(0 = y^2 - 3y\)

\(y = 0, 3\)

Now plug x back in.

\(x^2 - 2x = 0\)

\(\boxed{x = 0, 2}\)

\(x^2 - 2x = 3\)

\(\boxed{x = 3, -1}\)

Show that when the square of an odd integer is divided by 4, the remainder is always  =  1

Let  the  odd integer  =   2n -  1

(2n  - 1)^2 /  4  =   

( 4n^2 - 4n  +  1)  /   4  =

4n^2         4n              1

____  -  ______   +   ___   =

  4             4                4

n^2   -  n     +   1/4      →      remainder  = 1