# Extracurricular Elementary Mathematics Marathon (1 Viewer)

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Find with proof, all continuous functions f:R->R such that:

f(f(f(x)))=x for all real x.
I'm doubt the validity of my solution, and I'm certain there are massive logical gaps, but here's what I have:

$\bg_white \noindent Inverting the three-fold composition gives us f^{-1}(f^{-1}(f^{-1}(x)))=x \\ Thus, f(x) = f^{-1}(x), since f is arbitrary, and hence f(f(x)) = x. Substituting this back into the original functional equation gives f(x)=x, which obviously satisfies the original functional equation.$

#### seanieg89

##### Well-Known Member
I'm doubt the validity of my solution, and I'm certain there are massive logical gaps, but here's what I have:

$\bg_white \noindent Inverting the three-fold composition gives us f^{-1}(f^{-1}(f^{-1}(x)))=x \\ Thus, f(x) = f^{-1}(x), since f is arbitrary, and hence f(f(x)) = x. Substituting this back into the original functional equation gives f(x)=x, which obviously satisfies the original functional equation.$
Yeah, you have done something fishy in deducing that f=f^{-1}, care to explain your reasoning if you still believe this fact after thinking more?

Note also that you have used continuity nowhere. This is essential, as we have a vast array of solutions to the functional equation if continuity is not required:

Partition the reals into an uncountable union of sets, each with either 1 or 3 elements.
Define f to map elements of singleton sets to themselves and to cycle the three elements in each of the other sets.

Any such function f will satisfy the functional equation, but almost all of them will be highly discontinuous.

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$\bg_white \noindent f^3(x) = x \\ f^3(-x)=-x \\ Since composition preserves any existing parity (favouring even over odd) and cannot create parity, f(x) is odd. If f is a polynomial with degree m, then we have the following fact: O(f^3(x)) = O(x^{m^3}) = O(x) \Leftrightarrow m^3 = 1 \Leftrightarrow m=1 \\ Since f is odd, the polynomial f is f(x) = kx \\ Systematically bashing out the linear function yields f(x)=x \\ Not sure where to go from here...$

#### seanieg89

##### Well-Known Member
$\bg_white \noindent f^3(x) = x \\ f^3(-x)=-x \\ Since composition preserves any existing parity (favouring even over odd) and cannot create parity, f(x) is odd. If f is a polynomial with degree m, then we have the following fact: O(f^3(x)) = O(x^{m^3}) = O(x) \Leftrightarrow m^3 = 1 \Leftrightarrow m=1 \\ Since f is odd, the polynomial f is f(x) = kx \\ Systematically bashing out the linear function yields f(x)=x \\ Not sure where to go from here...$
Composition cannot create parity? Why not? Eg f(x)=1-x is not odd or even, but composed with itself is the identity which is odd.

It is true that the only polynomial function that works is the identity and yes degree considerations give you this quickly as you note.

This isn't too fruitful a way of thinking about what general continuous functions can solve the equation though.

##### -insert title here-
Composition cannot create parity? Why not? Eg f(x)=1-x is not odd or even, but composed with itself is the identity which is odd.

It is true that the only polynomial function that works is the identity and yes degree considerations give you this quickly as you note.

This isn't too fruitful a way of thinking about what general continuous functions can solve the equation though.
No discontinuities allowed, this includes singularities of rational functions? for example, 1/(1-x) is clearly a solution to the functional equation, but has a simple pole at x=1

#### seanieg89

##### Well-Known Member
No discontinuities allowed, this includes singularities of rational functions? for example, 1/(1-x) is clearly a solution to the functional equation, but has a simple pole at x=1
No, singularities are not allowed, because the function is continuous and has domain R. (We cannot have any domain "holes".)

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$\bg_white \noindent Fact 1: f is injective. \\ Proof: Suppose f is not injective, with x and y being distinct reals that satisfy f(x) = f(y). Then f(f(x)) = f(f(y)) and f(f(f(x))) = f(f(f(y))). But from the functional equation, f(f(f(x))) = x and f(f(f(y))) = y. So x=y, which contradicts the non-injectivity of f. Thus, f is injective. \\Fact 2: f is monotone. \\This follows from the requirement that f is entirely continuous in conjunction with Fact 1.\\ Condition 1: f(x) \leq x \\Proof: Suppose for some x, f(x)>x. So f(f(x)) > f(x) > x and f(f(f(x))) > f(f(x)) > f(x) > x. Then the functional equation is no longer true. Thus, f(x) \leq x \\ Condition 2: f(x) \geq x \\Proof: Suppose for some x, f(x)

$\bg_white \noindent Since both conditions must be simultaneously true, then the only such possibility is f(x) = x. Obviously, this satisfies the functional equation. \blacksquare\\ \\ Extension of this question:\\ Find all rational functions that satisfy the functional equation.$

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$\bg_white \noindent Show there are no solutions to the diophantine equation 2^n + 3^n = k^3 in positive integers (n,k)$

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$\bg_white Find all distinct integers n such that n+3^2 and n^2+3^3 are perfect cubes.$

$\bg_white \noindent Prove the following inequality: \\\sqrt{a^2 - \sqrt{3}ab+b^2}+\sqrt{b^2-bc+c^2} \geq \sqrt{c^2+a^2}$
$\bg_white \noindent Solve the following Diophantine Equation in \mathbb{Z} \\\\ a^3 + b^3 + c^3 = a^2 + b^2 + c^2$