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

 

Let f(x) = x^2− 2x. Find all real numbers x such that f(x) = f(f(x)).

 Feb 20, 2021
 #1
avatar+944 
+1

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

 

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

 

Let \(x^2 - 2x = y.\)

 

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

 

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

 

\(y = 0, 3\)

 

Now plug x back in.

 

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

 

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

 

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

 

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

 Feb 20, 2021
 #2
avatar+118069 
+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

 

 

cool cool cool

 Feb 20, 2021

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