Typos Give Rise To Problems

Algebra Level 3

Consider a function $f: \; \mathbb{Z} \to \mathbb{Z^*}$ such that $f(x) \cdot f(x+1) = f(x(x-1)).$

Evaluate $\dfrac{f(3)+f(1)}{3-1}$ .

3 0 2 1

1 solution

Guilherme Dela Corte
Dec 29, 2016

Knowing that $f(x) \cdot f(x+1) = f(x(x-1))$ , we can

plug $x = 0$ , and see that $f(0) \cdot f(1) = f(0) \Leftrightarrow f(1) = 1$ , since $f(0) \neq 0$ ;
plug $x = 2$ , and see that $f(2) \cdot f(3) = f(2) \Leftrightarrow f(3) = 1$ , since $f(2) \neq 0$ .

Thus, $\dfrac{f(3)+f(1)}{3-1} = \dfrac{2}{2} = \boxed{1.}$

Interesting setup. Can we characterize all such solutions?

Clearly, $f(x) = 1$ is a solution. Is it the only possible solution?

Calvin Lin Staff - 4 years, 5 months ago

I could deduce that $f(-1) = 1$ and found recurrences such as $f(0) = f(2), \; f(-2) = f(4) = f(6), \; f(12) = f(20)$ , ... Because so many equalities occur at some different values of $x$ in $f(x)$ , I could only guess that $f(x) = 1$ is the only solution.

However, it is curious that $f(x) = x(x-2), x \neq 0, 2$ is also a solution that yields $\dfrac{f(3)+f(1)}{3-1} = 1$ . This happens because $f(x+1) = f(x-1) \Leftrightarrow f(x) f(x+1) = {\color{#D61F06} f(x) f(x-1) = f(x(x-1))}$ .

I do not get why, in such cases, the red equation works. Also, what is an easy way to pull $f(x) = x(x-2)$ out of the hat?

Guilherme Dela Corte - 4 years, 5 months ago

What do you mean that you don't understand why the red equation works? Maybe it is clearer if you wrote it as

$f(x+1) = f(x-1) \Leftrightarrow f(x (x-1) ) = f(x) f(x+1) = f(x) f(x-1)$

I do not see a rigorous way to obtain $f(x) = x(x-2)$ (other than by guessing).

Calvin Lin Staff - 4 years, 5 months ago

Why f(0) and f(2) can not be eqau lto zero ??

Kushal Bose - 4 years, 5 months ago

@Kushal Bose – Because $f : \mathbb{Z} \to \mathbb{Z}^*$ , which means that $f$ can never be equal zero.

Guilherme Dela Corte - 4 years, 5 months ago

Typos Give Rise To Problems

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