Mathematical Reflections 2017, Issue 4 - Problem S416

Problem:
Let $f:\mathbb{N} \to \{\pm 1\}$ be a function such that $f(mn)=f(m)f(n)$ for all $m,n \in \mathbb{N}$. Prove that
there are infinitely many $n$ such that $f(n)=f(n+1)$.

Proposed by Oleksi Klurman, University College London, UK

Solution:
Let $n \in \mathbb{N}$. As $f$ is completely multiplicative, then $f(n^2)=(f(n))^2=1$. If there are infinitely many $n$ such that $f(n^2-1)=1$, we are done. So, assume that there are only finitely many $n$ such that $f(n^2-1)=1$. Then, there are infinitely many $n$ such that $f(n^2-1)=-1$. Since $f(n^2-1)=f(n-1)f(n+1)$, then there are infinitely many pairs $\{f(n-1),f(n+1)\}=\{-1,1\}$. Since $f(n)=\pm 1$, it follows that there are infinitely many $n$ such that $f(n)=f(n+1)$.

Mathematical Reflections 2017, Issue 4 - Problem S415

Problem:
Let $$f(x)=\dfrac{(2x-1)6^x}{2^{2x-1}+3^{2x-1}}.$$
Evaluate $$f\left(\dfrac{1}{2018}\right)+f\left(\dfrac{3}{2018}\right)+\ldots+f\left(\dfrac{2017}{2018}\right).$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Observe that $$f(1-x)=\dfrac{(1-2x)6^{1-x}}{2^{1-2x}+3^{1-2x}}=\dfrac{(1-2x)6^x}{2^{2x-1}+3^{2x-1}}=-f(x),$$ i.e. $f(x)+f(1-x)=0$.
So,
\begin{eqnarray*}
\sum_{k=1}^{1009} f\left(\dfrac{2k-1}{2018}\right)&=&\sum_{k=1}^{504}\left(f\left(\dfrac{2k-1}{2018}\right)+f\left(\dfrac{2018-(2k-1)}{2018}\right)\right)+f\left(\dfrac{1009}{2018}\right)\\&=&f\left(\dfrac{1009}{2018}\right)\\&=&f\left(\dfrac{1}{2}\right)\\&=&0.
\end{eqnarray*}

Mathematical Reflections 2017, Issue 4 - Problem J417

Problem:
Solve in positive real numbers the equation $$\dfrac{x^2+y^2}{1+xy}=\sqrt{2-\dfrac{1}{xy}}.$$

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Let $s=x+y$ and $p=xy$. Then, the given equation can be written as $$\dfrac{s^2-2p}{1+p}=\sqrt{2-\dfrac{1}{p}}.$$
By the AM-GM Inequality, we have $s^2-2p \geq 2p$, so $$\sqrt{2-\dfrac{1}{p}} \geq \dfrac{2p}{1+p} \iff \dfrac{(p-1)^2(2p+1)}{p(p+1)^2} \leq 0.$$ Since $p \geq 0$, it follows that $p=1$, which gives $s=2$. So, $x+y=2$ and $xy=1$, which yields $(x,y)=(1,1)$.

Mathematical Reflections 2017, Issue 4 - Problem J415

Problem:
Prove that for all real numbers $x,y,z$ at least one of the numbers
$$2^{3x-y}+2^{3x-z}-2^{y+z+1}$$ $$2^{3y-z}+2^{3y-x}-2^{z+x+1}$$ $$2^{3z-x}+2^{3z-y}-2^{x+y+1}$$
is nonnegative.

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Let $a=2^x$, $b=2^y$, $c=2^z$. Adding the three given numbers, we get
$$\begin{array}{lll} S&=&\left(\dfrac{a^3}{b}+\dfrac{a^3}{c}-2bc\right)+\left(\dfrac{b^3}{c}+\dfrac{b^3}{a}-2ca\right)+\left(\dfrac{c^3}{a}+\dfrac{c^3}{b}-2ab\right)\\&=&\left(\dfrac{a^3}{b}+\dfrac{b^3}{a}-2ab\right)+\left(\dfrac{b^3}{c}+\dfrac{c^3}{b}-2bc\right)+\left(\dfrac{c^3}{a}+\dfrac{a^3}{c}-2ca\right). \end{array}$$
By the AM-GM Inequality, we get
$$\dfrac{a^3}{b}+\dfrac{b^3}{a} \geq 2ab, \qquad \dfrac{b^3}{c}+\dfrac{c^3}{b} \geq 2bc, \qquad \dfrac{c^3}{a}+\dfrac{a^3}{c} \geq 2ca.$$
So, $S \geq 0$ and we conclude that at least one of the three given numbers is nonnegative.