SILBERBERG Gheorghe and a terrifying Problem : S.L.24.220

           The author, currently a teacher at UVT, was an international Olympian in mathematics. Originally from the city of Lugoj, he came with a bus full of students to the Timisoara County Olympics.

          Published in the Magazine(aka REVISTE) "GAZETA MATEMATICĂ : Supliment cu Exerciții", September 2024 at page 10. The statement is:

          "Let $f:\mathbb{R} \rightarrow \mathbb{Z}$ be a monotone and surjective function with properties:

              (i) $f(f(x))=f(x),\;\forall x\in \mathbb{R};$

              (ii) $f(2x)-f(x)=f\left ( x+\frac{1}{2}\right ),\; \forall x \in \mathbb{R}.$

     a) Show that $f\left ( k-\frac{1}{2^n} \right )=k-1$, whatever $k \in \mathbb{Z}$ and $n \in \mathbb{N}.$

     b) Determine the function $f$. "

ANSWER CiP

a) Induction on $n\in \mathbb{N}$

b) f(x)=[x] - the Floor function

                    Solution CiP

               The function $f$ being surjective, for $k\in\mathbb{Z}$ there exists $x_k\in\mathbb{R}$ such that $f(x_k)=k.$ Now $k=f(x_k)\underset{(i)}{=}f(f(x_k))=f(k)$, so
$$f(k)=k,\;\;\forall k\in\mathbb{Z} \tag{1}$$

          We see from (1) that the function $f$ is weakly increasing i.e. 

$$x<y\;\Rightarrow\;f(x)\leqslant f(y). \tag{1#}$$

          a) Equation

$$f\left (k-\frac{1}{2^n}\right )=k-1 \tag{2}$$

is true for $n=0$ according to (1). From (ii) we have

$$f\left (2\left (k-\frac{1}{2}\right)\right )-f\left (k-\frac{1}{2}\right )=f\left (k-\frac{1}{2}+\frac{1}{2}\right )\;\Leftrightarrow$$

$$\Leftrightarrow\;f(2k-1)-f\left (k-\frac{1}{2}\right )=f(k)\;\underset{(1)}{\Leftrightarrow}$$

$$\Leftrightarrow\;(2k-1)-f\left (k-\frac{1}{2}\right )=k\;\Rightarrow\;f\left (k-\frac{1}{2}\right )=k-1$$

so (2) is true for $n=1$.

{ edit nov 23, 2024: It seems that an inductive reasoning of (2) after $n$ has no immediate chance of success. } 

          From $(1)$ and $(1\#)$ we obtain the important estimate

$$k\leqslant x<k+1\;\;\;\Rightarrow\;k\leqslant f(x)\leqslant k+1,\;\;k\in\mathbb{Z}. \tag{3}$$

          Let us assume that for a certain $k_0\in \mathbb{Z}$ we have $\color {Red}{f\left (k_0-\frac{1}{4}\right ) \neq k_0-1}$. Hence, from (1#) follow $f\left ( k_0-\frac{1}{4}\right )=k_0$.

We have equality

$$f(2x+1)-f(2x)=f(x+1)-f(x) ,\;\;\forall x \in \mathbb{R}.\tag{4}$$

Indeed, applying (ii) to the underlined expressions

$$f(2x+1)-f(2x)=\underline{f \left ( 2 \left ( x+\frac{1}{2}\right ) \right )}-f(2x)=$$

$$=f\left (x+\frac{1}{2} \right )+f\left(x+\frac{1}{2}+\frac{1}{2} \right )-f(2x)=f(x+1)-\underline{f(2x)+f\left (x+\frac{1}{2} \right )}=$$

$$=f(x+1)-f(x).$$

Taking in (4) $x=k_0-\frac{1}{4}$ we get $f \left ( 2 \left (k_0-\frac{1}{4} \right )+1 \right )-f \left (2 \left( k_0-\frac{1}{4} \right ) \right )=f \left (k_0-\frac{1}{4}+1\right)-f(\left (k_0-\frac{1}{4} \right )\;\Leftrightarrow$

$$\Leftrightarrow\;f\left (2k_0+1-\frac{1}{2} \right )-f \left (2k_0-\frac{1}{2}\right )=f\left (k_0-\frac{1}{4}+1 \right )-f\left (k_0-\frac{1}{4} \right ) \;\Leftrightarrow$$

$$\overset{(2)\;for\;n=1}{\underset{f(k_0-1/4)=k_0}{\Leftrightarrow}}\;2k_0-(2k_0-1)=f\left (k_0+1-\frac{1}{4}\right )-k_0$$

hence $f\left (k_0+1-\frac{1}{4}\right )=k_0+1$, so applying a inductive reasoning results

$$f\left (k-\frac{1}{4}\right )=k,\;\forall k\geqslant k_0,\;k\in \mathbb{Z}.\tag{5}$$

Applying (ii) again for $k-\frac{1}{4}$ we get $f\left (2\left (k-\frac{1}{4}\right) \right )-f\left (k-\frac{1}{4} \right )=f\left (k-\frac{1}{4}+\frac{1}{2} \right )$

$$\Leftrightarrow\;f\left (2k-\frac{1}{2} \right )-f\left (k-\frac{1}{4} \right )=f\left ( k+\frac{1}{4} \right )\;\overset {(2)\;for\;n=1}{\underset{(5)}{\Leftrightarrow}}$$

$$\Leftrightarrow\;(2k-1)-k=f\left (k+\frac{1}{4} \right )\;\Leftrightarrow\;f\left (k+\frac{1}{4} \right )=k-1.$$

But then we get $k=f(k)\leqslant f \left ( k+\frac{1}{4} \right )=k-1$, FALSE.

 So $f\left (k-\frac{1}{4} \right )=k-1,\; \forall k \in \mathbb{Z}$   hence (2) is true for $n=2$.

{edit nov 27, 2024: This is how we can prove (2) by induction on $n$ }

          Let the predicate depending on the variable $n\in \mathbb{N}$ be

$$P(n)\;:\;\;"\forall k \in \mathbb{Z},\;f\left (k-\frac{1}{2^n} \right )=k-1"$$

     For $n\in \{0,\;1,\;2\}$ we have $P(n)-{\color{Green}{true}}$.

     We now assume $P(n)$-true for some $n$ and we will show that $P(n+1)$ is also true, so according to mathematical induction it results $\forall n\in \mathbb{N}\;P(n)$-true.

     If, by absurdity, $P(n+1)$ is false, it means that there exists $k_0\in \mathbb{Z}$ such that $f\left ( k_0-\frac{1}{2^{n+1}} \right ) \neq k_0-1$. But, because of (3), we must then have 

$$f\left (k_0-\frac{1}{2^{n+1}}\right )=k_0. \tag{6}$$

Applying (ii) for $x=k_0-\frac{1}{2^{n+1}}$ we get

$$f\left (2\left (k_0-\frac{1}{2^{n+1}}\right ) \right )-f\left ( k_0-\frac{1}{2^{n+1}}\right )=f\left ( k_0-\frac{1}{2^{n+1}}+\frac{1}{2} \right )\;\Leftrightarrow$$

$$\Leftrightarrow\;f\left ( 2k_0-\frac{1}{2^n}\right )-f \left (k_0-\frac{1}{2^{n+1}} \right )=f\left (k_0+\frac{1}{2}-\frac{1}{2^{n+1}} \right )\;\Leftrightarrow$$

$$\overset{P(n)-true}{\underset{(6)}{\Leftrightarrow}} \;(2k_0-1)-k_0=f\left (k_0+\frac{1}{2}-\frac{1}{2^{n+1}} \right )\;\;\Rightarrow$$

$$\Rightarrow\;\;\;f\left( k_0+\frac{1}{2}-\frac{1}{2^{n+1}} \right )=k_0-1.$$

But, since $\frac{1}{2}-\frac{1}{2^{n+1}}>0$, we have the inequalities 

$k_0=f(k_0)\leqslant f\left (k_0+\frac{1}{2}-\frac{1}{2^{n+1}} \right )=k_0-1$-FALSE.

          With this  "a)" is demonstrated.

          b) Let's show that for $k\leqslant x <k+1$ we have $f(x)=k$.

        If we choose a $n>-log_2(k+1-x)$ we have

 $-n<log_2(k+1-x)\;\;\Rightarrow\;\;2^{-n}<k+1-x\;\;\Rightarrow\;\;x<k+1-\frac{1}{2^n}.$

Further $k=f(k)\leqslant f(x)\leqslant f\left ( k+1-\frac{1}{2^n}  \right )\underset {(2)}{=}k$, so $f(x)=k$. We got the answer.

$\blacksquare$

A PROBLEM with the ISO_80000 SPECIFICATION

           "Find the real numbers $x$ and $y$, if  $\lg^2\frac{x}{y}=3\cdot \lg\frac{x}{2024}\cdot \lg \frac{2024}{y}$."

[Wikipedia says that, according to the ISO 80000 specification(it costs a lot to read it, and time...), "$\lg$" should be the standard notation for the decimal logarithm $log_{10}.$ ]

***  The sign "***" means that the problem does not have a known author, as it appears in the MAGAZINE(aka REVISTE) Supliment cu Exerciții, September 2024; proposed for the 10th grade on page 9 with number S.L24.211.

ANSWER CiP

$$x=2024,\;\;y=2024$$

Solution CiP

            Instead, we will solve the problem (...more than that, I don't venture to try...)
                    "Find the real numbers $x$ and $y$, if $\lg^2\frac{x}{y}=\lambda \cdot \lg\frac{x}{a} \cdot \lg \frac{a}{y}$"

       $a$ and $\lambda$ being given positive real numbers, $\color {Red}{\lambda <4}$."

                                                         The ANSWER  will be $\underline {x=a\;,\;y=a}$

     The existence conditions of the problem, $\frac{x}{y}>0,\; \frac{x}{a}>0,\;\frac{a}{y}>0$ combined give $x>0$ and $y>0$.

     I will use the inequality 

$$4\cdot u \cdot v \leqslant (u+v)^2 \tag{1}$$

with the sign $"="$ in (1) if and only if $u=v$. Let $A:=\lg\frac{x}{y}$;

$$A^2\underset{eq}{=}\lambda \cdot \lg\frac{x}{a}\cdot \lg\frac{a}{y}=\frac{\lambda}{4}\cdot 4\lg\frac{x}{a}\lg\frac{a}{y}\;\;\;\;\overset{(1)}{\underset{u=\lg\frac{x}{a}\;v=\lg\frac{a}{y}}{\leqslant}}\; \;\;\;\frac{\lambda}{4} \cdot \left ( \lg\frac{x}{a}+\lg \frac{a}{y} \right)^2 =$$

$$=\frac{\lambda}{4}\cdot \left [\lg \left (\frac{x}{a}\cdot \frac{a}{y}\right )\right ]^2=\frac{\lambda}{4}\cdot \lg^2 \frac{x}{y}=\frac{\lambda}{4}\cdot A^2,$$

hence $A^2 \leqslant \frac{\lambda}{4} \cdot A^2\;\Leftrightarrow\;A^2\cdot (1-\frac{\lambda}{4})\leqslant 0$ but which in the given condition $\lambda <4$ implies $A=0.$

     I got $\lg\frac{x}{y}=0$ so $x=y$ and, from the equation, that one of the conditions $\lg \frac{x}{a}=0$ or $\lg \frac{a}{y}=0$ occurs, so $x=a$ or $y=a$, (!)actually both.

$\blacksquare$

PROBLEM E:16993 author Mihaela BERINDEANU, Bucharest

 "Let $a,\;b\in \mathbb{N}^*$ be such that the number $\frac{a+3}{b}+\frac{b+3}{a}$ is an integer.

If $(a,b)$ is the greatest common divisor of the numbers $a$ and $b$, then show that $(a,b) \leqslant \sqrt{3(a+b)}$."

          From the MAGAZINE(aka REVISTE) Gazeta Matematica seria B no.9/2024, page 426, the proposed problem for the 7th grade

          ANSWER CiP

             A case where the equal sign occurs is

$$a=6,\;b=6.$$

          Solution CiP

          Let $k=\frac{a+3}{b}+\frac{b+3}{a}\in \mathbb{Z}$, actually $k\in\mathbb{N}^*$.

After calculations, we have the relationship

$$a^2+b^2-kab+3a+3b=0.\tag{1}$$

If $d=(a,b)$ then $a=d\cdot a_1,\;b=d \cdot b_1$ with $(a_1,b_1)=1.$Replacing these in (1) we have

$$d^2 \cdot a_1^2+d^2 \cdot b_1^2-k\cdot da_1\cdot db_1+3d\cdot a_1+3d\cdot b_1=0\;\Leftrightarrow$$

$$da_1^2+db_1^2-dka_1b_1+\underline{3(a_1+b_1)}=0.$$

In the last equation, the underlined term must be divisible by the number $d$, because all other terms are divisible by it. From here we get, with natural numbers

$$d \mid 3(a_1+b_1)\;\Rightarrow\;3(a_1+b_1)=d \cdot c \geqslant d\;\Rightarrow\;$$

$$\Rightarrow\;3\left ( \frac{a}{d}+\frac{b}{d} \right ) \geqslant d\;\Leftrightarrow \; 3(a+b)\geqslant d^2\;\;\Leftrightarrow\;(a,b)\leqslant \sqrt{3(a+b)}.$$

          For $a=6$ and $b=6$ we have $\frac{a+3}{b}+\frac{b+3}{a}=\frac{9}{6}+\frac{9}{6}=3$, and
$$d=(6,6)=6=\sqrt{36}=\sqrt{3\cdot(6+6)}.$$

 $\blacksquare$