Does there exist a function such that for every real number ?
This question belongs to the set Functions are awesome
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Suppose f exists. Let a = ϕ − 1 and b = − ϕ , where ϕ is the golden ratio 2 1 + 5 . Then f ( f ( a ) ) = a 2 − 2 = b and f ( f ( b ) ) = b 2 − 2 = a .
Consider the equation f ( f ( f ( f ( x ) ) ) ) = x . This simplifies to ( x 2 − 2 ) 2 − 2 = x , which is a polynomial equation of degree 4 . By the above paragraph, a and b are roots, and so are − 1 and 2 . So b , − 1 , a , 2 is the complete list of distinct real roots of the polynomial.
But f ( a ) is also a root, because f ( f ( f ( f ( f ( a ) ) ) ) ) = f ( a ) . So it must be one of the four numbers in the above list.
If f ( a ) = a then f ( f ( a ) ) = a = b , which is a contradiction.
If f ( a ) = b , then f ( b ) = f ( f ( a ) ) = b , and then f ( f ( b ) ) = f ( b ) = b = a , a contradiction.
If f ( a ) = − 1 , then b = f ( f ( a ) ) = f ( − 1 ) , but then a = f ( f ( b ) ) = f ( f ( f ( − 1 ) ) ) = f ( ( − 1 ) 2 − 2 ) = f ( − 1 ) , which is impossible since a = b .
The same argument shows f ( a ) = 2 is impossible.
So f cannot exist.
(This is a paraphrasing of a proof I saw in this Math Stack Exchange thread .)