joi, 9 ianuarie 2025

Andrika's Conjecture versus Rodika's Conjecture : On squares ending with three of the digit 4

           Andrica's conjecture is not a joke. Rodica's conjecture is a joke.

          We read in an article from the mathematics magazine "Revista Matematica a Elevilor din Timisoara", issue 1 of 1975: 

             "It is observed that the squares of some natural numbers end with identical digits (e.g. $12^2=144,\;1038^2=1\;077\;444$). 

     In the same issue of the magazine Dorin ANDRICA, then a diligent student from Deva, published some proposed problems, on pages 41-49: #2102, #2116, #2124, #2134, #2149.

     The cited article "Asupra unei proprietăți a numerelor naturale"(On a Property of Natural Numbers) with author N. I. NEDIȚĂ, on pages 3-6, explains the phenomenon. If a perfect square ends with dentical digits, these digits can only be 44 or 444. The general form of these numbers is, in the first case

$$50*k-38,\;\quad50*k-12,\quad \quad k=1,\;2,\;\dots \tag{1}$$

and in the second case

$$500*k-462,\;\quad500*k-38,\quad \quad k=1,\;2,\;\dots \tag{2}$$

There are no perfect squares that end with the group of digits 4444.


           Remark CiP  Formula (2) is included in Formula (1). To be more precise, certain values ​​of $k$ must be removed from (1).

 

          RODICA, an imaginary character, reads this article and makes the following Conjecture:

          "If a perfect square ends with three digits equal to 4, then the thousands digit of this square can only be odd."

     See the examples: $38^2=\color{Red} 1\;444,\;462^2=21\color{Red}3\;444,\;962^2=92\color{Red}5\;444$

$1038^2=1\;07\color{Red}7\;444,\;538^2=28\color{Red}9\;444$


duminică, 5 ianuarie 2025

'Neața NEAȚĂ ! (meaning "Good morning...")

          Ion NEAȚĂ from Slatina is also the author of Problem S:E24.271, published in the same issue of GMB, but this time in the Exercise Supplement (on page 7, proposed for 8th grade).

In translation:

  "S:E24.271  Knowing that $x,\;y\;$ and $A=\sqrt{4x^2+2y+6}+\sqrt{4y^2+2x+11}$ 

      are natural numbers, show that A is prime."

Answer CiP

$$x=1,\;\;y=3,\;\;\;A=11$$

Solution CiP

(obtained from a collaborator on AOPS)

          We first show the following:

              Lemma If  $a\;$ and $\;b\;$ are natural numbers s.t. $\sqrt{a}+\sqrt{b} \in \mathbb{N}\;$

                            then $a$ and $b$ are perfect squares.

               Proof (CiP) It is known that "the square root of $X$ is rational if and only if X is a rational number that can be represented as a ratio of two perfect squares" (see Wikipedia).

Now, if $\sqrt{a}+\sqrt{b}=m\in \mathbb{N}$ then $\sqrt{b}=m-\sqrt{a}$ and squaring it gives
$$\sqrt{a}=\frac{m^2+a-b}{2m}\in \mathbb{Q}$$

(the excluded case $m=0$ is trivialy). Hence $a$ is perfect square; but then $\sqrt{b}=m-\sqrt{a} \in \mathbb{Q}$ and so $b$ is a perfect square too.

$\square$ Lemma

          According to the Lemma, $A=\sqrt{4x^2+2y+6}+\sqrt{4y^2+2x+11} \in \mathbb{N}\;\Rightarrow$

$4x^2+2y+6=$ perfect square

$4y^2+2x+11=$ perfect square

so we have the inequalities (the first perfect square being an even number)

$$4x^2+2y+6 \geqslant (2x+2)^2,\;\;\;4y^2+2x+11 \geqslant (2y+1)^2$$

Adding the two inequalities

$$(4x^2+2y+6)+(4y^2+2x+11)\geqslant (4x^2+8x+4)+(4y^2+4y+1)\;\;\Leftrightarrow$$

$$\Leftrightarrow\;12\geqslant 6x+2y\;\;\;\Rightarrow \;\;\;\;y\leqslant 3(2-x). $$

Obvious $x \ngtr 2$. If $x=2\;\Rightarrow\;y=0$, but $4x^2+2y+6=22\neq$perfect square.

If $x=1\;\Rightarrow\;y\leqslant 3\;$ and $\;4x^2+2y+6=2y+10\leqslant 16$ it is a perfect square only for $y=3$.

If $x=0\;\Rightarrow\;y\leqslant 6\;$and $\;4x^2+2y+6=2y+6\leqslant 18$; it is possible only $y=5$ but then $4y^2+2x+11=100+11=11\neq$perfect square.

We got the answer.

$\blacksquare$

vineri, 3 ianuarie 2025

If we write QED it means it is GEOMETRY ??

 QED = GEOMETRY

More and more people are hesitant to use the expression QED at the end of a demonstration.


          Regarding problem E:17017 from GMB 10/2024.

In translation:
               "E:17017. Let $ABC$ be any triangle with $AB=3 \cdot BC$. Consider the 
 points $D\in AC$ such that $AD= 3 \cdot DC$, and $E$ the midpoint of the side $AB$.
 Show that $BD \perp DE$."

Solution CiP

          Due to the dimensions given in the problem it is convenient to choose $DC=p$ and $AB=6\cdot m$, hence $AD=3 \cdot p,\;BE=AE=3\cdot m,\;BC=2 \cdot m$. 

          It is obvious that $\frac{AE}{AC'}=\frac{3m}{4m}=\frac{3}{4}=\frac{3p}{4p}=\frac{AD}{AC}$ so with the Reciprocal of Thales' Theorem
$$DE \parallel CC' \tag{1}$$
          It is also obvious that $\frac{BC}{BA}=\frac{2m}{6m}=\frac{1}{3}=\frac{p}{3p}=\frac{CD}{AD}$ so with the Converse of Angle Bisector Theorem $BD$ is the bisector of angle $\measuredangle CBA$.

Then, in the obvious isosceles triangle $BCC'$ the bisector of angle at vertex $B$ is also the altitude, so $BD \perp CC'$. From (1) it then follows $BD \perp DE$.

QED

$\blacksquare$


            Remark CiP

          A more computational solution uses the Law of Cosines.

     In $\Delta ABC\;: \quad cos\; C=\frac{CB^2+CA^2-AB^2}{2\cdot CB \cdot CA}=\frac{4m^2+16p^2-36m^2}{2\cdot 2m\cdot 4p}=\frac{p^2-2m^2}{mp}$

and in $\Delta BCD\;:\quad BD^2=CB^2+CD^2-2\cdot CB \cdot CD \cdot cos\;C=$

$=4m^2+p^2-2 \cdot 2m \cdot p \cdot \frac{p^2-2m^2}{mp}=4m^2+p^2-4(p^2-2m^2)$ so

$$BD^2=12m^2-3p^2 \tag {2}$$

Again in $\Delta ABC\;:\quad cos\;A=\frac{AB^2+AC^2-BC^2}{2\cdot AB \cdot AC}=\frac{36m^2+16p^2-4m^2}{2\cdot 6m \cdot 4p}=\frac{p^2+2m^2}{3mp}$

and in $\Delta ADE\;: \quad DE^2=AD^2+AE^2-2\cdot AD \cdot  AE \cdot cos\;A=$

$=9p^2+9m^2-2\cdot 3p \cdot 3m \cdot \frac{p^2+2m^2}{3mp}=9p^2+9m^2-6(p^2+2m^2)$ so

$$DE^2=3p^2-3m^2 \tag{3}$$

From (2) and (3) it result

$$BD^2+DE^2=(12m^2-3p^2)+(3p^2-3m^2)=9m^2=BE^2$$

hence by Converse of Pythagorean Theorem the angle between sides $BD$ and $DE$ is a right angle.

<end Rem>


miercuri, 25 decembrie 2024

Ce? se întâmplă când are toată familia febră ?? // What happens when the whole family has a fever?

          It seems that the Mathematical Olympiad has begun.

          Let's try to put ourselves in the shoes of a 5th grader. Let's take a list of issues given not too long ago. For example, let's consult the magazine below. 

(Unfortunately I can't show you more than that.)
- The picture is courtesy of SSMR -

          We will choose three 5th grade problems from the first stage of the Olympiad as examples.


          Problem 1 (at page 1) "A test has 20 questions, which are to be solved in 150

                       minutes. The questions are of three types: easy, medium and difficult. 

                       It is expected that each of the 7 easy questions can be solved in 4 

                      minutes and each of the medium questions can be solved in 8 minutes.

                       How many minutes are left for solving difficult questions?"      

                               A. 56            B. 58            C.62            D. 66            E. 70


Answer CiP        D. 66

Solution CiP

               Easy questions and medium questions require a solving time of $7 \cdot 4+7\cdot 8=24$  minutes. So for the difficult questions, $150-84=66$ minutes remain. The answer is D.

$\blacksquare$

               Remark CiP We have 7 easy questions and 7 medium questions, so the number of difficult questions is $20-7-7=6$. So each difficult question is expected to be solved in 11 minutes.

< end Rem>


          In light of the statement in Problem 1, it seems that Problems 1-7 are considered easy, Problems 8-14 are medium, and Problems 15-20 are difficult. I will take one of each. 

           I also mention that the Magazine in the image above does not contain the corresponding answers to these questions. Despite the other qualities: e.g. a clear English that would make even Shakespeare envious. News about the adventures at this Olympics also appeared in the press. That's also where I found out where you can find the Answers. The same lovely SSMR.


          Problem 11 (at page 2) "The largest integer which divided by 2022 gives a

                    quotient smaller than the remainder is :

A. 1048043482         B. 4086461         C. 16185         D. 8091         E. 9         " 

 

Answer CiP    B. 4086461

Solution CiP

               By dividing a number $D$ by $2022$ we obtain the quotient $Q$ and the remainder $R$, according to the long division as below

$$\begin{array}{c|c} D & 2022 \\ \hline \vdots & Q \\ \hline R & \; \end{array}$$

and the Division with remainder Theorem is written

$$D=2022\cdot Q+R\;\; \quad R<2022 \tag{1}$$

The statement also specifies the condition 

$$Q\;<\;R \tag{2}$$

So the first relation (1) gives us $D<2022R+R$ , or $D<2023R$. But from the second relation (1) we  have $D\leqslant 2021$, and so

$$D<2023 \cdot 2021 =4\;088\; 483.$$

So the answer "A."  is excluded, being a larger number.

     We see from (1) that the largest admissible number $D$ occurs when $R$ and $Q$ are the largest, also verifying the condition (2), so 

$$Q=R-1=2020.$$

Hence $D=2022 \cdot 2020+2021=4\;086\;461$, so answer is B

$\blacksquare$

           Remark CiP  Answer options C and E also verify the condition (2)

$$16\;185 \;:\;2022 \;=\;8\quad remainder\;9\;\quad \quad 9\;:\;2022\;=\;0\quad remainer\;9$$

but option D does not $8\;091\;;\;2022\;=\;4\quad remainder\;3.$

<end Rem>


          Problem 18 (at page 3) "If $S(n)$ denotes the sum of the digits of the number

                  $n$, then the number of $n-$s so that $\;2021\leqslant n+S(n)\leqslant 2022$ is:

A. 0         B. 1         C. 2         D. 3         E. 4    "


Answer CiP   C. 2

$$1996+S(1996)=1996+25=2021\;\quad \; 2014+S(2014)=2014+7=2021$$

Solution CiP

          $n=0$ does not satisfy the condition.

           If $n\leqslant 1999$ we have $S(n)\leqslant 28$, and then

$$n\geqslant 2021-S(n)\geqslant 2021-28=1993.$$

We have to try the numbers:

$$1993,\quad 1994,\quad\cdots,1999.\quad  \tag{1}$$

After some calculations:

$$1993+S(1993)=1993+22=2015,$$

$$1994+S(1994)=1994+23=2017,$$

$$1995+S(1995)=1995+24=2019,$$

$1996+S(1996)=1996+25=2021$ - this number is convenient

we see that for the other numbers in the list (1) the value $n+S(n)$ exceeds 2023.

          If $n\geqslant 2000$ (anyway $n\leqslant 2022$) we have $S(n)\geqslant 2$ so

 $n\leqslant 2022-S(n)\leqslant 2020$.

Among the numbers   2000, 2001, ..2020 the highest $S(n)$ value is $S(2019)=12$ and then, from $n\geqslant 2021-S(n)$ we get $n\geqslant 2021-12=2009$, so we have to try only the numbers

$$2009,\quad 2010,\;\dots\;2020.\quad \tag{2}$$

Observing the evolution of calculations:

$$2009+S(2009)=2009+11=2020$$

$$2010+S(2010)=2010+3=2013$$

$$2011+S(2011)=2011+4=2015$$

$$2012+S(2012)=2012+5=2017$$

$$2013+S(2013)=2013+6=2019$$

$2014+S(2014)=2014+7=2021$ - this number is convenient

$$2015+S(2015)=2015+8=2023$$

it turns out that from here on out, there can be no more convenient numbers in list (2).

$\blacksquare$

luni, 18 noiembrie 2024

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.

$\blacksquare$




miercuri, 6 noiembrie 2024

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.

$\blacksquare$

marți, 5 noiembrie 2024

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
$$d=(6,6)=6=\sqrt{36}=\sqrt{3\cdot(6+6)}.$$

 $\blacksquare$