Does the function exists?

Algebra Level 4

Does there exist a function f : R R f : \mathbb{R}→\mathbb{R} such that f ( f ( x ) ) = x 2 2 f(f(x)) = x^2-2 for every real number x x ?

This question belongs to the set Functions are awesome

No Yes

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1 solution

Patrick Corn
Jan 30, 2018

Suppose f f exists. Let a = ϕ 1 a = \phi-1 and b = ϕ , b = -\phi, where ϕ \phi is the golden ratio 1 + 5 2 . \frac{1+\sqrt{5}}2. Then f ( f ( a ) ) = a 2 2 = b f(f(a)) = a^2-2 = b and f ( f ( b ) ) = b 2 2 = a . f(f(b)) = b^2-2 = a.

Consider the equation f ( f ( f ( f ( x ) ) ) ) = x . f(f(f(f(x)))) = x. This simplifies to ( x 2 2 ) 2 2 = x , (x^2-2)^2-2 = x, which is a polynomial equation of degree 4. 4. By the above paragraph, a a and b b are roots, and so are 1 -1 and 2. 2. So b , 1 , a , 2 b,-1,a,2 is the complete list of distinct real roots of the polynomial.

But f ( a ) f(a) is also a root, because f ( f ( f ( f ( f ( a ) ) ) ) ) = f ( a ) . 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 f(a)=a then f ( f ( a ) ) = a b , f(f(a))=a \ne b, which is a contradiction.

If f ( a ) = b , f(a)=b, then f ( b ) = f ( f ( a ) ) = b , f(b)=f(f(a))=b, and then f ( f ( b ) ) = f ( b ) = b a , f(f(b))=f(b)=b \ne a, a contradiction.

If f ( a ) = 1 , f(a)=-1, then b = f ( f ( a ) ) = f ( 1 ) , b=f(f(a))=f(-1), but then a = f ( f ( b ) ) = f ( f ( f ( 1 ) ) ) = f ( ( 1 ) 2 2 ) = f ( 1 ) , a=f(f(b))=f(f(f(-1)))=f((-1)^2-2)=f(-1), which is impossible since a b . a \ne b.

The same argument shows f ( a ) = 2 f(a)=2 is impossible.

So f f cannot exist.

(This is a paraphrasing of a proof I saw in this Math Stack Exchange thread .)

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