# Prove f(f(f(x))) `!=` f(x) in ZxZ xy+2x+y=1

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### 1 Answer

You need to find f(x) = y.

Factoring by y to the left side yields:

`y(x+1)+ 2x = 1`

Isolating y in the given relation yields:

`y = (1 - 2x)/(x + 1)`

Composing the function f with itself yields:

`fof = f(f(x))`

Replacing x by y yields`f(f(x)) = f((1 - 2x)/(x + 1)) = f(y)` `f((1 - 2x)/(x + 1)) = (1 - 2f(y))/(f(y)+1)`

Composing the function f(f(x)) with f yields:

`f(f(x))of =f(f(f(x))) = f((1 - 2f(y))/(f(y)+1)) = (1-2(1 - 2f(y))/(f(y)+1))/((1 - 2f(y))/(f(y)+1) + 1)`

`f(f(f(x))) = (1 - (2- 4f(y))/(f(y)+1))/((1 - f(y) + 1)/(f(y) + 1))` =>`f(f(f(x))) = -(3f(y)+ 1)/(2 - f(y))`

`f(f(f(x))) = - ((3-6x)/(x+ 1) + 1)/(2 - (1 - 2x)/(x + 1))`

f(f(f(x))) = -(4 - 5x)/1

**Notice that f(f(f(x))) `!=` f(x).**