(Difficult) How can I prove such function doesn't exist?

Question

Show there's no function f:R --> R such that the composite h = f o f is given by h(x) = x^2 - 1996. This problem is considered difficult. I don't know how to do this.

Answer 1

I can prove that no CONTINUOUS f can satisfy this equation.
First note that h(x) has two fixed points, 1/2+1/2*sqrt(7985) and 1/2-1/2*sqrt(7985). Next note that if y is a fixed point of h, then so is f(y):
h(f(y))=f(f(f(y)))=f(h(y))
=f(y).
I can show that f(1/2+1/2*sqrt(7985))
=1/2-1/2*sqrt(7985) and f(1/2-1/2*sqrt(7985))
=1/2+1/2*sqrt(7985).
Next, note that since
f(1/2-1/2*sqrt(7985))>0
and
f(1/2+1/2*sqrt(7985))<0,
then
f(1/2-1/2*sqrt(7985))
-(1/2-1/2*sqrt(7985))>0
and
f(1/2+1/2*sqrt(7985))
-(1/2+1/2*sqrt(7985))<0,
the continuous function f(x)-x must have a zero. Let this zero be z, so that f(z)-z=0.
Then
f(z)=z so z is a fixed point of f. Thus
z^2-1996=h(z)=f(f(z))=f(z)=z so z=1/2-1/2*sqrt(7985) or 1/2+1/2*sqrt(7985). Since f(z)=z, we must have
f(1/2-1/2*sqrt(7985))
=1/2-1/2*sqrt(7985)
or
f(1/2+1/2*sqrt(7985))
=1/2+1/2*sqrt(7985).
But we already know that neither of these is true, so this contradiction shows that no continuous f can satisfy h(x)=f(f(x)).

Answer 2

Ksoileau's answer was great. However, I think there's a proof that ensures such function doesn't exist without any additional assumption, like continuity.

Suppose there exists such f according to the statement and let g = f o f. Let's try to find 2 distinct real numbers a and b such that g(a) = b e g(b) = a (such points are known by a name I don't remember. It's something that reminds oscillation, I guess. If you know, please, email me). This leads us to the following system:

a^2 - 1996 = b
b^2 - 1996 = a

Subtracting the 2nd equation from the 1st and keeping in mind a and be are distinct, we get (a+b)(a-b) = (b-a), so that a = - b - 1. Now, plugging this in the 1st equation we get 2 distinct real roots. Their values are of no interest to us, what matters is they are distinct, since we clearly notice we won't get a = b = -1/2. So, what really counts is that we came to the following conclusion: If that f exists, then there exists only one pair (a, b) with the property that g sends a into b and b into a. When I say only one pair, I'm taking into account that, for our purpose, (a, b) and (b ,a) can be considered, without loss of generality, as the same pair, since it doesn't matter if a comes first or if it's b that comes first.

Now, let c = f(a) and d = f(b). Applying f to both sides of each equality, we get b = f(c) and a = f(d). Applying again, gives f(b) = g(c) and f(a) = g(d). This implies that d = g(c) and c = g(d). So, the pair (c, d) has the same property as that of the pair (a, b). But, since we saw before that there was only one pair with such property, we must have a = c, and b = d or a = d and b = c.

If a = c and b = d, then we get a = f(d) = f(b) = g(c) = g(a) = b. But this contradicts the previous conclusion that and b are distinct.

If a = d and b = c, then a = f(d) = f(a) = g(d) = c = b, again a contradiction. .

So, in any case, the assumption that there exists f such that f(f(x) = x^2 - 1996 leads to a contradiction, and we finally conclude such function doesn't exist.

THE END (thanks God!)

Answer 3

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Answer 4

Hint: Use the vertical line test