ISL 2003 A5

by Wolstenholme, Aug 4, 2014, 11:55 PM

Let $\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ that satisfy the following conditions:

- $f(xyz)+f(x)+f(y)+f(z)=f(\sqrt{xy})f(\sqrt{yz})f(\sqrt{zx})$ for all $x,y,z\in\mathbb{R}^+$ ;

- $f(x)<f(y)$ for all $1\le x<y$ .

Solution:

Lemma 1: $f(x) \geq 2$ for all $x \in \mathbb{R}^{+}$ .

Proof: Note that by letting $x = y = z = 1$ in the equation we find that $f(1) = 2$ and then letting $y = z = 1$ in the equation and replacing $x$ by $x^2$ we have that $f(x^2) = f(x)^2 - 2$ . This implies $f(x) > \sqrt{2}$ for all $x \in \mathbb{R}^{+}$ . Using this result we now get $f(x) > \sqrt{2+\sqrt{2}}$ for all $x \in \mathbb{R}^{+}$ . Continuing in this fashion we obtain the desired result.

Lemma 2: Let $n$ be a nonnegative integer. Let $p_{n}$ be a polynomial defined by the following criteria: $p_0 = 2, p_1 = x, p_{k + 1} = xp_k - p_{k - 1}$ for all positive integers $k$ . Then $f(t^n) = p_n(f(t))$ for all $t \in \mathbb{R}^{+}$ .

Proof: It is clear that $p_n\left(t + \frac{1}{t}\right) = t^n + \frac{1}{t^n}$ for all real $t$ (this is easily proven by induction) so it suffices to show that $f(t^{2n + 1}) = f(t)f(t^{2n}) - f(t^{2n - 1})$ and $f(t^{2n}) = f(t^n)^2 - 2$ . We already know that $f(x^2) = f(x)^2 - 2$ for all $x \in \mathbb{R}^{+}$ so the second identity follows immediately. To prove the first identity, let $x = t^{2n -1 }$ and $y = z = t$ in out equation. Then we obtain $f(t^{2n + 1}) + f(t^{2n - 1}) + 2f(t) = f(t)f(t^{n})^2 = f(t)(f(t^{2n}) + 2)$ and so $f(t^{2n + 1}) = f(t)f(t^{2n}) - f(t^{2n - 1})$ as desired.

Lemma 1 implies that for any $s \in \mathbb{R}^{+}$ we can write $f(s) = a + \frac{1}{a}$ for some real $a \geq 1$ . Lemma 2 then implies that $f(s^q) = a^q + \frac{1}{a^q}$ for all $q \in \mathbb{Q}$ . Because the rationals are dense in the reals and because $f$ is increasing on the interval $[1, \infty)$ this implies that $f(s^r) = a^r + \frac{1}{a^r}$ for all $r \in \mathbb{R}$ . This clearly implies that $f(x) = x^m + \frac{1}{x^m}$ for some $m \in \mathbb{R}$ and it is easy to check that all such functions satisfy the original equation.

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glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
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by fukano_2, Jun 14, 2020, 6:24 AM
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by AopsUser101, Jan 29, 2020, 8:27 PM
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by mathleticguyyy, Jun 5, 2019, 8:26 PM
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by User360702, Jan 10, 2019, 6:03 PM
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by songssari, Jun 12, 2016, 8:21 AM
shouts make blogs happier
by briantix, Mar 18, 2016, 9:57 PM
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by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

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ISL 2003 A5

by Wolstenholme, Aug 4, 2014, 11:55 PM

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