An UNSOLVED completely problem from Sierpiński

 It is Problem #4, page 16 mentioned in the cited work. Edited from the manuscript.

In translation : 

                           "Prove that there are infinitely many natural numbers  $n$  for which

                          the number  $4n^2+1$ is divisible by both  $5$  and  $13$."

               The author's[WS] solution is on page 38. (The text is written in green.) In translation:

                         "ANSWER : All numbers in the arithmetic progression

 $65k+56,\;\;k=0,\;1,\;2\dots$

[I noticed at the bottom of the page in the first photo that there is no indication of how to find the answer. Then comes its Solution :]

          Indeed, if  $n=65k+56$ , $k\geqslant 0$ is an integer, then  $n\equiv 1\;(mod\;5)$ and $n\equiv 4\;(mod\;13)$ from where  $4n^2+1\equiv 0\;(mod\;5)$ and  $4n^2+1\equiv 0\;(mod\;13)$, so that 

$5\mid 4n^2+1$  and  $13\mid 4n^2+1.$

$\color {Green}{\blacksquare}$"

                I solved the problem without knowing this answer. [Text written in blue; in translation:]

ANSWER CiP

The numbers that have the property in the statement are exactly those of the form

$65k+4,\;\;65k+9,\;\;65k+56,\;\;65k+61$  where  $k=0,\;1,\;2,\dots\;.$

                   Solution CiP

               Since  $5$  and  $13$  are coprime, we have

$5\mid A\;\;and\;\;13 \mid A\;\;\;\Leftrightarrow\;\;\;5\cdot 13 \mid A$

Let  $n=65k+r$ ;  then  $4n^2+1=4(65k+r)^2+1=4(65^2k^2+2\cdot 65\cdot r+r^2)+1=$

$=65\cdot(4\cdot 65k^2+8r)+4r^2+1=65k_1+4r^2+1.$

Trying to choose a number  $r$  so that  $4r^2+1=65$  we get  $r^2=16$. For example  $r=4$, so we have an infinity of numbers, of the form  $n=65k+4$, with the property in the statement. The problem would be solved [, but not completely.]

$\color {Blue}{\blacksquare}$

       I noted in red, further on, that we can also have  $r=-4$, obtaining another infinity of convenient numbers. 

         - And all the convenient values ​​found in my answer were obtained by checking all the possibilities

$65k,\;65k\pm1,\;65k\pm2,\;\dots,\;65k\pm32$

     Finally, I also noted, [written in black], that : The numbers in ANSWER CiP form the sequence

 A203464 in OEIS ("The On-Line Encyclopedia of Integer Sequences).

АXA! Немам ПОИМ!! // AHA! I have NO IDEA!!

 It's a parody of the title of Martin ERICKSON's  book "Aha! Solutions" 

          The Problems Given at the Dam for JBMO in North Macedonia fell into my hands. I said I'd try my hand at Problem 1. 

            "1.  Let  $n>1$ be a natural number and  $m>2$  a divisor of  $2n$. Prove that

               the number  $n^2$ can be written as the sum of  $m$ nonzero perfect squares."

          I have no idea how to solve this, so for now I'm just showing a few

          EXAMPLES CiP

           a) $n=2$ :  $2<m \mid 4\;\;\Rightarrow\;\;m=4$.  We have the equation

$2^2=1+1+1+1$

          b) $n=3$ :  $2<m \mid 6\;\;\Rightarrow\;\;m=3$ or $m=6$. We have the equations

$3^2=1+4+4$

$3^2=1+1+1+1+1+4$

          c) $n=4$ :  $2<m \mid 8\;\;\Rightarrow\;\;m=4$ or $m=8$. We have the equations

$4^2=4+4+4+4$

$4^2=1+1+1+1+1+1+1+9$

Let's hope that inspiration will help me solve the problem.

Edited Friday 06 June  The following Lemma, which I will now formulate only as a Conjecture, would be helpful:

               Lemma CiP (Conjecture) Whatever the prime number  $p>2$, the

                                                  number  $p^2$ can be written as the sum of  $p$ squares.

          So we have an equation like this

$$p^2=a_1^2+a_2^2+\dots +a_p^2=\sum_{i=1}^pa_i^2 \tag{P}$$

          Examples:

$3^2=\underset{3-terms}{\underbrace{1+4+4}}$

$5^2=\underset{5-terms}{\underbrace{4+4+4+9}}$

$7^2=\underset{7-terms}{\underbrace{1+1+4+9+9+9+16}}$

For the following example we proceed by trial and error:

$11^2=\underset{4-terms}{\underbrace{5+16+36+64}}\;\overset{we\;replace\; 16\; with\; a}{\underset{sum\;of\;several\;terms}{=}}\;\underset{7-terms}{\underbrace{5+4+4+4+4+36+64}}=$

and now if we replace the term  $5$ with the sum  $1+1+1+1+1$  we are lucky to obtain the desired result, so

$11^2=\underset{11-terms}{\underbrace{1+1+1+1+1+4+4+4+4+36+64}}$

Until we find a proof for the Lemma, let us observe that if for two prime numbers $p>2$ and $q>2$ we have decompositions  (P) and

$$q^2=\sum_{j=1}^qb_j^2$$

then because

$$\left ( \sum_{i=1}^pa_i^2\right )\cdot \left (\sum_{j=1}^qb_j^2 \right )=\sum_{i,j=1}^{i=p,j=q}a_i^2b_j^2$$

we obtain a sum of  $p\cdot q$  squares 

$$p^2\cdot q^2=\sum_{i,j=1}^{i=p,j=q}(a_i\cdot b_j)^2 \tag{PQ}$$

<end Edit 06 June>

Weekend Edition (There are no days off in Mathematics. When you are struggling with a problem, you are constantly thinking about it.) 

          But what if the property in yesterday's Lemma holds for any number  $n>2$ ? 

          Also by trying, as for number  $121$ , I obtained the following equations:

$4^2=\underset{4-terms}{\underbrace{4+4+4+4}}$

$6^2=\underset{6-terms}{\underbrace{1+1+1+4+4+25}}$

$8^2=\underset{8-terms}{\underbrace{1+1+1+1+1+1+9+49}}$

$9^2=\underset{9-terms}{\underbrace{1+1+1+1+1+4+4+4+64}}$

Ugh!, no pattern is visible. Worse, if we start from equation for  $3^2=1+4+4$ and square it (or rather multiply it by itself),

$3^2 \cdot 3^2=(1+4+4)\cdot (1+4+4)=\dots $ (the $3\times 3=9$ terms obtained by multiplication are all squares)

 we find an equation for  $9^2$ that differs from what I obtained:

$9^2=\underset{9-terms}{\underbrace{1+4+4+4+4+16+16+16+16}}$

So, we don't have unique writings for the representations we're looking for. Complicated stuff...

<end Weekend Edition>

          I couldn't be patient anymore and I consulted the solution to the problem. As expected, the problem should be quite simple. I'm saved !

          You can also consult the solution from where I got the problem statement.

          As a consolation for how much I've been struggling these days, the statement that I considered above as a Lemma is true. In fact, the following are true:

    (a)   For any number  $n\geqslant 3$, the number  $n^2$ can be written as a sum

            of  $n$ nonzero perfect squares.

    (b)   For any number  $n\geqslant 2$, the number  $n^2$ can be written as a sum

             of  $2\cdot n$ nonzero perfect squares.

 To my shame, these statements are almost trivial. If you only showed them, you would get 2 points on the solution scale. For just one, you would get nothing.

        Proof of (a) : We have that

  $n^2=(n^2-4\cdot n +4)+4\cdot n-4=(n-2)^2+4\cdot (n-1)$  so

$$n^2=\underset{1-term}{\underbrace{(n-2)^2}}+\underset{n-1\;-terms}{\underbrace{4+4+\dots+4}}$$

     Examples (that differ from those I found) : 

$6^2=4^2+4+4+4+4+4$

$7^2=5^2+4+4+4+4+4+4$

$8^2=6^2+4+4+4+4+4+4+4$

$9^2=7^2+4+4+4+4+4+4+4+4$

$10^2=8^2+4+4+4+4+4+4+4+4+4$

$11^2=9^2+4+4+4+4+4+4+4+4+4+4$

       Proof of (b) :  We have that

$n^2=(n^2-2\cdot n+1)+2n-1=(n-1)^2+(2n-1)\cdot 1$   so

$$n^2=\underset{1-term}{\underbrace{(n-1)^2}}+\underset{2\cdot n-1\;-terms}{\underbrace{1+1+\dots +1}}$$

Note that (a) is the particular case with  $m=n$ of the Problem, and (b) is the particular case  $m=2n$ of it.

                    Official Solution (adapted by CiP)

Let  $k=\frac{2n}{m}$; it is, from the divisibility condition, a natural number, $k\geqslant 1$. From

$n^2=(n^2-2 n k+k^2)+2nk-k^2\;\;\overset{2n=k\cdot m}{=}\;(n-k)^2+km\cdot k-k^2=(n-k)^2+k^2 \cdot (m-1)$

and because  $n-k=\frac{2n}{2}-k=\frac{k\cdot m}{2}-k=k\cdot \left (\frac{m}{2}-1\right )>0$

we have

$$n^2=(n-k)^2+\underset{m-1\;-terms}{\underbrace{k^2+\dots +k^2}}$$

QED $\blacksquare$

मलाही या समस्येची लाज वाटेल.

   शीर्षकाचे भाषांतर असे आहे: I would be ashamed of this problem too"

Someone, signed "ANONYMOUS", commented on yesterday's post. I wanted to delete the comment, usually insulting, but I researched it and the guy is right. He, in the comment, refers to this problem:

In translation:

                      "10.     Show that if   $a,\;b,\;c \in \mathbb{R}$  and  $ab+bc+ca=0$ , then

$2\sqrt{a^2+b^2+c^2}\geqslant 3\sqrt[3]{|abc|}.$

C. Ionescu-Țiu "

Indeed, the problem is a bit strange. Unless it is a typographical error, it insults the intelligence of a solver with minimal knowledge. I won't bother with it any further, I'll just say this:

 it is true under any conditions for any numbers  $a,\;b,\;c$  not just those that satisfy the relation $ab+bc+ca=0$. In addition, the  $=$  sign only occurs in the case of  $a=b=c=0$.

          Proof  CiP  For non-negative numbers $x,\;y,\;z$ , holds the inequality

  AM-GM :   $\frac{x+y+z}{3}\geqslant \sqrt[3]{xyz}$ 

 so

$x+y+z\geqslant 3\cdot \sqrt[3]{xyz}$

Replacing above $x=a^2,\;y=b^2,\;z=c^2$, where the numbers  $a,\;b,\;c$  are arbitrary, not bound by any condition, we obtain

$a^2+b^2+c^2\geqslant 3\cdot \sqrt[3]{a^2b^2c^2}$

and taking the square root of both sides we have

$$2\cdot \sqrt{a^2+b^2+c^2}\geqslant 2\sqrt{3}\cdot \sqrt[3]{|abc|}$$

But, since  $2\sqrt{3}=\sqrt{12}>\sqrt{9}=3$, the above results in

$$2\sqrt{a^2+b^2+c^2}>3\cdot \sqrt[3]{|abc|}\;.$$

$\blacksquare$

More in Joke, more in Serious : A Problem Close to Logic

 I don't expect C. Ionescu-Țiu to show much logic in His Problems. Here is Problem E:6061 from the magazine in the picture.

I have published more about this issue of the Magazine elsewhere.

In translation:

                        "E:6061*. Consider the real and positive numbers  $a,\;b,\;c,\;d$  such

                        that $a+b=c+d$.  Show that:

                          1). If  $ab>cd$  then  $a^2+b^2<c^2+d^2$  and the converse.

                          2). If  $ab>cd$  then  $|a-b|<|c-d|$.

                          3). If  $a^2+b^2<c^2+d^2$  then  $|a-b|<|c-d|$  and the converse."

          Solution CiP (an improvised solution, at the school level)

               Let us remember that everywhere in what follows holds the equation:

$a+b=c+d \tag{1}$

               1). Direct implication:         $ab>cd\;\Rightarrow\;a^2+b^2<c^2+d^2$

(1)$\;\Rightarrow\;\;(a+b)^2=(c+d)^2$

$\Rightarrow\;\;a^2+b^2+2\cdot ab=c^2+d^2+2\cdot cd$

$\Rightarrow\;\;a^2+b^2-c^2-d^2=2\cdot (cd-ab)$

and from hypothesis  $ab>cd$  it follows  $cd-ab<0$,  so  $a^2+b^2-c^2-d^2<0$, or equivalently : $a^2+b^2<c^2+d^2$.

qed

                    Converse implication :          $a^2+b^2<c^2+d^2\;\Rightarrow\;ab>cd$ 

              $a^2+b^2<c^2+d^2\;\;\Rightarrow\;\;-a^2-b^2>-c^2-d^2\;\;\underset{(1)}{\Rightarrow}$

$\Rightarrow\;\;(a+b)^2-a^2-b^2>(c+d)^2-c^2-d^2\;\;\Leftrightarrow\;\;2\cdot ab>2\cdot cd\;\;\Leftrightarrow\;\;ab>cd$

qed

          Remark CiP  Based on the equation

$a^2+b^2-c^2-d^2=2\cdot (cd-ab)$

we have the logical equivalence

$a^2+b^2-c^2-d^2<0\;\;\;\Leftrightarrow\;\;\;cd-ab<0$

and the statement  "1)" is obtained immediately.

<end Rem>

               2). Implication:   $ab>cd\;\;\Rightarrow\;\;|a-b|<|c-d|$

$ab>cd\;\;\Rightarrow\;\;-4\cdot ab<-4\cdot cd\;\;\Rightarrow\;\;(a-b)^2-(a+b)^2<(c-d)^2-(c+d)^2\;\;\Rightarrow$

$\;\underset{(1)}{\Rightarrow}\;\;(a-b)^2<(c-d)^2\;\;\Rightarrow\;\;\sqrt{(a-b)^2}<\sqrt{(c-d)^2}\;\;\Leftrightarrow\;\;|a-b|<|c-d|$.

qed

          Remark CiP

                 a)  The converse is also valid:  $|a-b|<|c-d|\;\;\Rightarrow\;\;ab>cd$

Because  $|a-b|<c-d|\;\;\Rightarrow\;\;(a-b)^2<(c-d)^2\;\;\Rightarrow\;\;-(a-b)^2>-(c-d)^2\;\;\Rightarrow$

$\underset{(1)}{\Rightarrow}\;\;(a+b)^2-(a-b)^2>(c+d)^2-(c-d)^2\;\;\Leftrightarrow\;\;4\cdot ab>4\cdot cd$, &c...

                 b) As in the Remark from point 1), we can prove point 2) together with its converse, based on the equation

$(a-b)^2-(c-d)^2=4\cdot(cd-ab)$

<end Rem>

               3).  Direct implication:     $a^2+b^2<c^2+d^2\;\;\Rightarrow\;\;|a-b|<|c-d|$

From  $2a^2+2b^2<2c^2+2d^2\;\;\Rightarrow\;\;(a+b)^2+(a-b)^2<(c+d)^2+(c+d)^2\;\;\Rightarrow$

$\underset{(1)}{\Rightarrow}\;\;(a-b)^2<(c-d)^2\;\;\Rightarrow\;\;\sqrt{(a-b)^2}<\sqrt{(c-d)^2}\;\;\Leftrightarrow\;\;|a-b|<|c-d|$

qed

                     Converse implication:     $|a-b|<|c-d|\;\;\Rightarrow\;\;a^2+b^2<c^2+d^2$

From the hypothesis  $|a-b|<|c-d|$  follows immediately the inequality  

$(a-b)^2<(c-d)^2 \tag{2}$

$\Rightarrow\;\;2a^2+2b^2-4ab<2c^2+2d^2-4cd\;\;\Rightarrow\;\;2(a^2+b^2-c^2-^2)<4ab-4cd=$

$=(a+b)^2-(a-b)^2-[(c+d)^2-(c-d)^2]\underset{(1)}{=}(c-d)^2-(a-b)^2\underset{(2)}{<}0.$

qed

               Remark CiP   Both implications, direct and converse, result at once from the equation

$2(a^2+b^2-c^2-d^2)=(c-d)^2-(a-b)^2$

<end Rem>

With this we solved the exercise, and something extra.

                    Final Remarks CiP

               1.  For the logical propositions:

$p :  ab>cd$

$q :  a^2+b^2<c^2+d^2$

$r :   |a-b|<|c-d|$

we have the logical equivalents

$p\;\;\Leftrightarrow\;\;q\;\;\Leftrightarrow\;\;r$

               2.  Due to the symmetry in $a$ and $b$ on the one hand and in $c$ and $d$ on the other hand, we could assume from the beginning that  $a\leqslant b$ and $c\leqslant d$. With this, the proposition $p$ is true  $\Leftrightarrow\;\;ab\underset{(1)}{>}c(a+b-c)\;\;\Leftrightarrow\;\;c^2-c(a+b)+ab>0\;\;\Leftrightarrow\;\;(c-a)(c-b)>0\;\;\Leftrightarrow\;\;c\in (0,\;a)\cup (b,\;+\infty)$.

and similar for $d$, so ultimately we have on the real axis the ordering $0<c<a\leqslant b<d$, to which is added the condition that the segments with ends $a$ and $b$, respectively $c$ and $d$ have the same midpoint.

$\blacksquare$

Nicușor DAN - A Day in the Life of a President // Nicușor DAN - O Zi din Viața unui Președinte

 Your doctoral thesis is inaccessible to the public. In its place we put a work from our youth...

His work is not too extensive either...

Diophantine Equation $x^2+y^2=2\cdot z^2$ in terms of the correct solution of the Pythagorean Equation

 In this post we will solve in integers the equation

$$x^2+y^2=2\cdot z^2 \tag{E}$$

Noting that  $(x,y,z)=(\pm a,\pm b,\pm c)$ are solutions, once one of them is, we will limit ourselves to solutions in positive integers. 

We will see that the solution of equation (E) depends on the general solution of the Pythagorean equation $x^2+y^2=z^2$. But I have explained this in more detail here.

          If the integers $(x,y,z)$ verify the equation (E) then, since  $2\cdot z^2$ is an even number, $x$ and  $y$  must necessarily have the same parity. So $\frac{x+y}{2}$ and  $\frac{x-y}{2}$ are integers. It can be seen that (E) is equivalent to

$$\left ( \frac{x+y}{2}\right )^2+\left ( \frac{x-y}{2} \right )^2=z^2 \tag{P}$$

which shows that  $\left (\frac{x+y}{2},\frac{x-y}{2},z \right )$  are solutions of the Pythagorean equation.

     Returning to the previously mentioned Post, the positive integers solutions (not only the primitive ones) of the equation  (P) are given by the families with parameters $s,\; t$ and $d$ ($d$ being a multiplicity factor):

$\begin {cases}\frac{x+y}{2}=2dst\\\frac{x-y}{2}=d(s^2-t^2) \tag{1}\\z=d(s^2+t^2) \end{cases}$

and

$\begin{cases}\frac{x+y}{2}=d(s^2-t^2)\\\frac{x-y}{2}=2dst \tag{2}\\z=d(s^2+t^2)\end{cases}$

where $d\in \mathbb{Z}$ and

$s>t>0$ are two coprime integers of opposite parity.                        (st)

          Then the general solution in positive integers of equation (E) will be, solving for (1) and (2):

$x=d\cdot (s^2+2st-t^2),\;\;y=d\cdot (-s^2+2st+t^2),\;\;z=d\cdot (s^2+t^2) \tag{3.1}$

and

$x=d\cdot (s^2+2st-t^2),\;\;y=d\cdot (s^2-2st-t^2),\;\;z=d\cdot (s^2+t^2) \tag{3.2}$

We notice that the value of  $x$  in the two formulas is the same. And the value of  $y$  is given by two formulas differing only in sign: for $s>\sqrt{2}\cdot t$  we have $y_1<0$ and for $s<\sqrt{2}\cdot t$  we have  $y_2<0$. Since we are only interested in positive values, we can express the final result in the form

          The integer and positive solutions of the equation (E) are

$x=d\cdot (r^2+2st-t^2),\;\;y=d\cdot |s^2-2st-t^2|,\;\;z=d\cdot (s^2+t^2) \tag{DST}$

where $d\in \mathbb{N}$  and  $s,\;t$  check the condition (st) above.

          Remark CiP  In the table below we show some solutions for $d=1$ (i.e. primitive solutions, gcd(x,y,z)=1)

ATTENTION! The values ​​$s=9,\; t=7$  do not satisfy the condition (st). The solution $(x,y,z)=(158,94,130)=2\cdot (79,47,65)$ is the multiple of another solution...

Karmaşık görünen 29 071 numaralı sorunun çok basit olduğu ortaya çıktı // The Problem 29 071 that seemed complicated turned out to be too trivial

The author, Mihály Bencze, is famous, which is perhaps why the Magazine GM-B accepted this issue.

In tranlation

                          "29071.   Let the function  $f\;:\;\mathbb{N} \to \mathbb{N}$  be defined by

$f(n)=\left [ \frac{n}{3}\right ]+\left [ \frac{n+1}{5} \right ]+\left [ \frac{n+2}{7}\right ]$ ,  for all $n\in\mathbb{N}.$

                                  Prove that the function is neither injective nor surjective."

ANSWER CiP

$f(0)=f(1)=0$  , so the function is NOT injective;

$f(8)=4,f(9)=6$ and the function $f$, which is increasing, never takes the value $5.$

                Solution CiP

               It was not said in the statement, as many other times, that  $[\alpha]$  denotes the integer part of the real number  $\alpha$. By definition, $[\alpha]\in\mathbb{Z}$  is the (uniquely determined) integer that satisfies one of the conditions below:

$$[\alpha]\leqslant \alpha <[\alpha]+1,\quad \quad \alpha -1<[\alpha]\leqslant \alpha\;. \tag{][}$$

 

          The case of injectivity:

$f(0)=\left [ \frac{0}{3}\right ]+\left [ \frac{1}{5}\right ]+\left [\frac{2}{7}\right ]=0+0+0=0:$

$f(1)=\left [\frac{1}{3}\right ]+\left [\frac{2}{5}\right ]+\left [\frac{3}{7}\right ]=0+0+0=0.$

So the function $f$ is NOT injective.

          The case of surjectivity:

     Let's first note that $f$ is increasing, because function $[x]$ is: $x<y\Rightarrow [x]\leqslant [y]$

        (The pedantic solver would do this:

        let $x<y;$ if $x\leqslant [y]$ then, because $[x]\leqslant x$ we deduce from the transitivity of the inequalities that $[x]\leqslant [y];$

                           if $[y]<x$ then, because $x<y\overset{(][)}{<}[y]+1$ we deduce from transitivity that $[y]<x<[y]+1\overset{(][)}{\Rightarrow} [x]=[y]$.)

          We see that $f(8)=\left [2\frac{2}{3}\right ]+\left [1\frac{4}{5}\right ]+\left [1\frac{3}{7}\right ]=2+1+1=4$,

$f(9)=\left [\frac{9}{3}\right ]+\left [\frac{10}{5} \right ]+\left [1\frac{4}{7}\right ]=3+2+1=6$

so $f(n)$ takes the values ​​4 and 6 for the consecutive numbers 8 and 9, so, being increasing, it cannot take the value 5 for any natural number $n$. (The pedant would say so: $n\leqslant 8\Rightarrow f(n)\leqslant 4;\;\;n\geqslant 9\Rightarrow f(n)\geqslant 6$  and there are no other possibilities.  )

The function $f$ is not surjective.

$\blacksquare$

            Remark CiP

          I suspected from the beginning that the values ​​of the function $f$ make "jumps". I looked for values ​​of $n$ for which each of the three fractions would be a natural number. For this we set the conditions:

$$\begin {cases}n=3\cdot k\;\;\;\;\;\;\;\;(1)\\n+1=5\cdot l\;\;\;(2)\\n+2=7\cdot m\;\;(3)\end{cases}$$

We substitute (1) into (2) and we obtain

$$3\cdot k+1=5\cdot l\;\;\Leftrightarrow\;\;5\cdot l-3\cdot k=1 \tag{kl}$$ 

which is a linear Diophantine equation. We see the solution $k=3,\;l=2$ of (kl) so it has the general solution

$$k=5\cdot p+3,\;\;l=3\cdot p+2 \;,\;\;p\in\mathbb{N}\tag{4}$$

Then  $n\underset{(1)}{=}3\cdot k\underset{(4)}{=}3(5p+3)=15\cdot p+9$ which substituted into (3) gives us

$$(15\cdot p+9)+2=7\cdot m\;\;\Leftrightarrow\;\;7\cdot m-15\cdot p=11 \tag{mp}$$

Here we see an solution  $m=-7,\;p=-4$, so the general solution of  (mp) is

$$m=15\cdot t-7,\;\;p=7\cdot t-4\;,\;\;t\in\mathbb{N} \tag{5}$$

Finally  $n=15p+9\underset{(5)}{=}15(7\cdot t-4)+9=105\cdot t-51.$

     The first value of  $n$ we are looking for is $n=105-51=54.$ Calculating, we get

$f(53)=\left [\frac{53}{3}\right ]+\left [ \frac{54}{5}\right ]+\left [\frac{55}{7}\right ]=\left [17\frac{2}{3}\right ]+\left [10\frac{4}{5}\right ]+\left [ 7\frac{6}{7}\right ]=17+10+7=34,$

$f(54)=\left [\frac{54}{3}\right ]+\left [\frac{55}{5}\right ]+\left [\frac{56}{7}\right ]=18+11+8=37,$

$f(55)=\left [\frac{55}{3}\right ]+\left [\frac{56}{5}\right ]+\left [\frac{57}{7}\right]=\left[18\frac{1}{3}\right]+\left [11\frac{1}{5}\right]+\left[8\frac{1}{7}\right]=18+11+8=37.$

From here it is immediately clear that the function  $f$  is neither injective nor surjective.

<end Rem>

          A PEDANT's remark  Instead of (1)-(3) we could use congruences according to different moduli. 

$n\equiv 0\;(mod\;3),\;\;n+1\equiv 0\;(mod\;5)\Leftrightarrow n\equiv  -1\;(mod\;5),\;\;n\equiv -2\;(mod\;7).$

 Now,  $3\cdot k \equiv -1\;(mod\;5)\Rightarrow 6\cdot k \equiv -2\;(mod\;5)\Leftrightarrow k\equiv -2\;(mod\;5).$ So  $k=5\cdot p-2,\;p\in\mathbb{Z}$  and further  $n=3\cdot k=15\cdot p-6\equiv -2\;(mod\;7)\Rightarrow 15\cdot p\equiv 4\;(mod\;7)\Rightarrow p\equiv 4\equiv -3\;(mod\;7).$ Thus  $p=7\cdot t-3,\;t\in\mathbb{Z}$, so $n=15p-6=15(7t-3)-6=105\cdot t-51.$

<end rem>

İkisi Bir Arada: Yağ ve su gibi karışmazlar // Two in One : they don't mix, like oil and water

 Strange juxtaposition of two issues in SGMB 3/2025

In translation:

     "Let  $a,\;b,\;c$  be positive real numbers. Show that

$\frac{\sqrt{a}}{b+c}=\frac{\sqrt{b}}{a+c}=\frac{\sqrt{c}}{a+b} \tag{E}$

                                  if and only if  $a=b=c.$"

                                        Solution CiP

            Let's assume that  $a\neq b$. Then, from the first two of the equalities (E), we get

$0<\frac{\sqrt{a}}{b+c}=\frac{\sqrt{b}}{a+c}=\frac{\sqrt{a}-\sqrt{b}}{(b+c)-(a+c)}=\frac{\sqrt{a}-\sqrt{b}}{b-a}=-\frac{\sqrt{a}-\sqrt{b}}{a-b}=$

$=-\frac{\sqrt{a}-\sqrt{b}}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=-\frac{1}{\sqrt{a}+\sqrt{b}}<0$

but this is ABSURD. 

          So $a=b$, and therefore from the equalities (E) there remains

$$\frac{\sqrt{a}}{a+c}=\frac{\sqrt{c}}{2\cdot a}$$

Assuming now that  $c\neq a$, the above results in

$0<\frac{\sqrt{a}}{a+c}=\frac{\sqrt{a}-\sqrt{c}}{(a+c)-2a}=-\frac{\sqrt{a}-\sqrt{c}}{a-c}=-\frac{1}{\sqrt{a}+\sqrt{c}}<0$

this is again ABSURD. So  $c=a=b$.

QED $\blacksquare$

Pell's Equations $\;\;x^2-D\cdot y^2=\pm1$

           Let Peel's Equation be

$$x^2-D\cdot y^2=1.\tag{P}$$

where $D$ is a given positive nonsquare integer.

     In solving (P) we must find the fundamental solution, i.e. a pair of positive integers $(x_1,y_1)$ wich satisfie them,and $x_1$ is minimal. An algorithm that determines this, found by me in the book "Dr. Paul RADOVICI-MĂRCULESCU, Probleme de Teoria Elementară a Numerelor" (Ed. Tehnică, București, 1986), at page 178, uses Continued Fraction.

      We develop $\sqrt{D}$ in continued fraction (their shape is well known):

$\sqrt{D}=[a_0,\; \overline{a_1,\;\dots,\;a_{k-1},\;2a_0}] \tag{D}$

                  $\blacklozenge$ if $k$=odd, $k=2t+1$ then the fundamental solution is  $x_1=A_{4t+1},\;y_1=B_{4t+1}; \tag{O}$

                   $\blacklozenge$ if $k$=even, $k=2t$ then the fundamental solution is $x_1=A_{2t-1},\;y_1=B_{2t-1}.\tag{E}$

I wrote $A_k$ the numerator and $B_k$ the denominator of the $k$th convergent of (D), i.e.

$[a_0,\;a_1,\;\dots, a_k]=\frac{A_k}{B_k}.$

I found their calculation formulas in the book ""I.M. VINOGRADOV, Bazele Teoriei Numerelor"(Ed. Academiei, 1957), at page 17

$$A_i=a_i\cdot A_{i-1}+A_{i-2},\;\;B_i=a_i\cdot B_{i-1}+B_{i-2},\;\;i\geqslant 1$$

with initial values  $A_{-1}=1,\;B_{-1}=0;\,\,\,A_0=a_0,\;B_0=1$

where A_n is the numerator and B_n is the denominator, called continuants,^[1][2] of the nth convergent.

       Schematically, obtaining $A_i$ can be represented as follows

$\begin{cases}\;\;\;\;\;\;\;\;\times \swarrow a_i\\\underset{+}{\underbrace{A_{i-2}\;A_{i-1}}}\;\fbox{Ai}\tag{S}\end{cases}$

And similar for $B_i$.

          Example 1        $x^2-13\cdot y^2=1$

         We have the continued fraction expansion

$\sqrt{13}=[3,\;\overline{1,1,1,1,6}]$   so $k=5$.

We apply the scheme (S), the calculated elements being written in color, and we obtain the table:

\begin{array}{|c|c}\hline\\\;i&-1&0&1&2&3&4&5&6&7&8&\color{Green}9&10\\\hline \\a_i&\;&3&1&1&1&1&6&1&1&1&1&6\\\hline \\A_i&1&3&\color{Red}4&\color{Red}7&\color{Red}{11}&\color{Red}{18}&\color{Red}{119}&\color{Red}{137}&\color{Red}{256}&\color{Red}{393}&\color{Green}{649}&\color{Red}{4287}\\\hline\\B_i&0&1&\color{Red}1&\color{Red}2&\color{Red}3&\color{Red}5&\color{Red}{33}&\color{Red}{38}&\color{Red}{71}&\color{Red}{109}&\color{Green}{180}&\color{Red}{1189}\\\hline \end{array}

We are in the case (O),  $k=5=2\cdot t+1\;\Rightarrow t=2$ and the minimal solution is (colored green)

$x_1=A_9=649,\;\;y_1=B_9=180.$

          Remark CiP  The result is in agreement with that given in the Table on page 127 of the book

ANDREESCU Titu, ANDRICA Dorin, CUCUREZEANU Ion - An Introduction to Diophantine Equations : A Problem-Based Approach (Birkhäuser Springer, New York Dordrecht Heidelberg London, 2010)

<end Rem>

          Example 2         $x^2-31\cdot y^2=1$

          We have the continued fraction expansion

$\sqrt{31}=[5,\;\overline{1,1,3,5,3,1,1,10}]$    so  $k=8.$

Now we are in the case (E):  $8=k=2\cdot t\;\Rightarrow\;t=4$  and the minimal solution is, according to the calculations in the table below

$x_1=A_7=1520,\;\;y_1=B_7=273.$

(Compare with the Table mentioned in the previous Remark.)

\begin{array}{|c|c|}\hline\\i&-1&0&1&2&3&4&5&6&\color{Green}7\\\hline \\a_i&\;&5&1&1&3&5&3&1&\color{Green}1\\\hline\\A_i&1&5&6&11&39&206&657&863&\color{Green}{1520}\\\hline\\B_i&0&1&1&2&7&37&118&155&\color{Green}{273}\\\hline\end{array}

          As for ALL the solutions in natural numbers of the equation (P), they are given by the formulas (O) - wiyh $t=0,\,1,\,2,\dots$ and (E) - with $t=1,\;2,\dots$. However, the further calculation of $n$th convergent  is too laborious.

          A simpler approach involves writing a system of recurrence relations.

          Suppose, for the equation (P) that we have found the fundamental solution $(x_1,y_1)$. Then, using the matrix

$P=P_D=\begin{pmatrix}x_1&D\cdot y_1\\y_1&x_1\\\end{pmatrix} \tag{M}$

we write the linear recurrence relation in the form 

$\begin{pmatrix}x_{n+1}\\y_{n+1} \end{pmatrix}=P \cdot \begin{pmatrix}x_n\\y_n \end{pmatrix}=\begin{pmatrix}x_1&D\cdot y_1\\y_1&x_1 \end{pmatrix} \cdot \begin{pmatrix}x_n\\y_n \end{pmatrix}\;,n=0,\;1,\dots\tag{MX}$

with  $(x_0,y_0)=(1,0)$ which is the universal solution for the equations (P),

that is

$\begin{cases}x_{n+1}=x_1\cdot x_n+(Dy_1)\cdot y_n\\y_{n+1}=y_1\cdot x_n+x_1 \cdot y_n \end{cases},\;\;x_0=1,\;y_0=0,\;\;\;n=0,\;1,\dots \tag{XY}$

     The two-recurrence system (XY) can be transformed into second-order recurrence relations, as shown in another post.

        Example 3             $x^2-3\cdot y^2=1$

        Fundamental solution of this equation is $(x_1,y_1)=(2,1)$. So  $P=\begin{pmatrix}2&3\cdot 1\\1&2\end{pmatrix}$  and the recurrence relations are written

$\begin{cases}x_{n+1}=2\cdot x_n+3\cdot y_n\\y_{n+1}=x_n+2 \cdot y_n\end{cases},\;x_0=1,\;y_0=0\;\;\;(n=0,\;1,\dots).$

The first few solutions appear in the Table below:

\begin{array}{c|c}n&0&1&2&3&4&5&6\\\hline \\x_n&1&2&7&26&97&362&1351\\\hline \\y_n&0&1&4&15&56&209&780 \end{array}

       Or we can write the second-order recurrence relations, as mentioned above, with the same results:

$x_{n+2}=4\cdot x_{n+1}-x_n,\;\;x_0=1,\;x_1=2\;\;\;\;;\;\;\;\;y_{n+2}=4\cdot y_{n+1}-y_n,\;\;y_0=0,\;y_1=1\;\;(n=0,\;1,\dots)$.

          Now let the Negative Pell (or "minus Pell") equation be 

$x^2-D\cdot y^2=-1 \tag{-P}$

Not all of these equations have solutions. But when they do, they solve similarly. I first encountered it in an article in GM-P, no. 4/1989, pages 175-178: Gheorghe UDREA - "Asupra Formei Solutiilor Ecuatiilor  $x^2-D\cdot y^2=\pm1$". We quote from here what interests us.

         Let's go back to (D).

$\quad\blacklozenge$ if $k$=odd the solutions of the equation (-P) are

$x_n=A_{k(2n-1)-1},\;\;\;y_n=B_{k(2n-1)-1},\;\;\;n=1,\;2,\;3,\dots ;\tag{-O}$

$\quad\blacklozenge$ if $k$=even the equation (-P) has NO solutions.

So, the smallest positive solution, when it exists (so $k$ is odd) is

$x_1=A_{k-1}\,,\,\,\,y_1=B_{k-1} \tag{-F}$

Furthermore, we can also determine the smallest positive solution(the fundamental one) of the equation (P) by the formula 

$x_1'+y_1'=(x_1+y_1\cdot \sqrt{D})^2 \tag{+P}$

          Example 4    $x^2-13\cdot y^2=-1$

         We have (see Example 1) $\sqrt{13}=[3,\overline{1,1,1,1,6}]$  so $k=5$. Minimal solution is $x_1=A_4=18,\;y_1=B_4=5.$ Further we have

$(18+5\cdot \sqrt{13})^2=324+25\cdot 13+180\cdot \sqrt{13}=649+180 \cdot \sqrt{13}$

so minimai solution of $x^2-13\cdot y^2=1$ is $x_1'=649,\;y_1'=180$ as we have already found.

Images of the older manuscript and some drafts.

The Lost Notebooks - I : Transforming a System of Linear Recurrences of Order 1 into (two) Linear Recurrences of Order 2

          Let the sequences  $(x_n)_{n\in \mathbb{Z}},\;(y_n)_{n\in \mathbb{Z}}$  be defined by the relations 

\begin{cases}x_{n+1}=a\cdot x_n+b\cdot y_n\;\;\;\;(1)\\y_{n+1}=c \cdot x_n+d \cdot y_n\;\;\;\;\;(2) \end{cases}

To determine them, an "initial" value needs to be given, for example $x_0$ and  $y_0$.

          We show that we can write a recurrence relation for each of the sequences in (1), (2), surprisingly (or NOT ?) the same.

$$x_{n+2}=(a+d)\cdot x_{n+1}-(ad-bc)\cdot x_n\;\;;\;\;y_{n+2}=(a+d)\cdot y_{n+1}-(ad-bc)\cdot y_n \tag{3}$$

To determine exactly each of the sequences $(x_n)_{n\in \mathbb{Z}},\;(y_n)_{n\in\mathbb{Z}}$, two "initial" values ​​need to be given, for example $x_0$ and $x_1$ &c.

                       Proof  CiP (with the necessary clumsiness)

               $x_{n+2}\;\overset{(1)}{\underset{\overbrace{n\to n+1}}{=}}\;a\cdot x_{n+1}+b\cdot y_{n+1}\overset{(2)}{=}\;a\cdot x_{n+1}+b\cdot (          c\cdot x_n+d\cdot y_n)=$

$=a\cdot x_{n+1}+bc\cdot x_n+bd\cdot y_n\;$

$\Rightarrow \;\;\;bd \cdot y_n=x_{n+2}-a\cdot x_{n+1}-bc\cdot x_n \tag{4}$

Later $d\cdot x_{n+1}\overset{(1)}{=}ad\cdot x_n+bd\cdot y_n \overset{(4)}{=}ad\cdot x_n+x_{n+2}-a\cdot x_{n+1}-bc\cdot x_n\;\;\Rightarrow$

$\Rightarrow\;\;\;x_{n+2}=(a+d)\cdot x_{n+1}-(ad-bc)\cdot x_n\;;$ the first formula (3) is proven. 

     From (2) written for $n+1$, it follows

$y_{n+2}-d\cdot y_{n+1}=c\cdot x_{n+1}\overset{(1)}{=}c\cdot(a\cdot x_n+b\cdot y_n)\;\overset{\underbrace{c\cdot x_n=from\;(2)}}{=}a\cdot (y_{n+1}-d\cdot y_n)+bc\cdot y_n=$

$=a\cdot y_{n+1}-(ad-bc)\cdot y_n$

and from the extreme equalities the second part of (3) results.

$\blacksquare$

          Remark CiP  The matrix $M=\begin{pmatrix}a&b\\c&d\\\end{pmatrix}$ that appears in (1)&(2) has the characteristic polynomial $P(\lambda)=\lambda^2-(a+d)\cdot \lambda+(ad-bc)$, which coincides with the characteristic polynomial of the recurrences (3).

<end Rem>

          An example is shown in the images below. (In the second image, with the red color, I made the mistake of eliminating $x_n$ and obtained for $y_n$ a recurrence relation of order 3 - in formula (4) there. This turned out to be just an illusion...)

A Pell's equation hidden in Problem 29 052

 In translation:

                     "29052.   Let $(x_n)_{n\geqslant 1}$ be the string with $x_n=[(3+2\sqrt{2})^n],\;n\geqslant 1$.

                   Prove that for any $n\in \mathbb{N}^*$, the number  $\frac{(x_n-1)(x_n+3)}{8}$  is a perfect square.

                      ($[\alpha]$ represents the integer part of the real number $\alpha$.)

Mihai Pîrvulescu, Calafat"

ANSWER CiP

$$\frac{(x_n-1)(x_n+3)}{8}=B_n^2 \tag{1}$$

                         where $B_n$ is defined by  

$B_1=2,\;B_2=12,\;\;B_{n+2}=6\cdot B_{n+1}-B_n,\;n\geqslant 1. \tag{2}$

                    Solution CiP

               Let

$(3+2\sqrt{2})^n=A_n+B_n\cdot \sqrt{2},\;n\geqslant 1. \tag{3}$

We have $A_1=3,\;B_1=2$, and since $(3+2\sqrt{2})^2=17+12\sqrt{2}$, we get $A_2=17,\;B_2=12.$ Examining the expansion with Binomial formula of $(3\pm2\sqrt{2})^n$, it results that

$(3-2\sqrt{2})^n=A_n-B_n\sqrt {2}. \tag{4}$

Moreover

$(3+2\sqrt{2})^n+(3-2\sqrt{2})^n\;\;\overset{(3),(4)}{=}\;2\cdot A_n. \tag{5}$

          Writing $(3+2\sqrt{2})^{n+1}=(3+2\sqrt{2})^n\cdot(3+2\sqrt{2})$, we have from (3)

$A_{n+1}+\sqrt{2}B_{n+1}=(A_n+\sqrt{2}B_n) \cdot (3+2\sqrt{2})=(3A_n+4B_n)+\sqrt{2}(2A_n+3B_n),$ so

\begin{cases} A_{n+1}=3A_n+4B_n\;\;\;\;\;\;\;\;\;\;\;\;(6) \\B_{n+1}=2A_n+3B_n\;\;\;\;\;\;\;\;\;\;\;\;(7)\end{cases}

Using relations (6), (7) several times we have

$B_{n+2}\;\overset{(7)}{\underset{n\to n+1}{=}}\;2A_{n+1}+3B_{n+1}\overset{(6)}{=}2(3A_n+4B_n)+3B_{n+1}=3B_{n+1}+8B_n+6A_n=$

$\overset{(7)}{\underset{\overbrace{2A_n=B_{n+1}-3B_n}}{=}}\;\;3B_{n+1}+8B_n+3(B_{n+1}-3B_n)=6B_{n+1}-B_n.$

So the sequence $(B_n)_{n \geqslant 1}$ checks the relations (2).

          Let us now prove that $(x,y)=(A_n,B_n)$ verifies Pell's equation

$x^2-2y^2=1. \tag{8}$

 Obviously $(A_1,B_1)=(3,2)$ verifies (8). Then assuming

$A_n^2-2B_n^2=1 \tag{9}$

we have $A_{n+1}^2-2B_{n+1}^2\overset{(6)}{\underset{(7)}{=}}(3A_n+4B_n)^2-2(2A_n+3B_n)^2=(9A_n^2+24A_nB_n+16B_n^2)-$

$-2(4A_n^2+12A_nB_n+9B_n^2)=A_n^2-2B_n^2\overset{(9)}{=}1,$

so (9) is also true for  $n+1$, so by the Principle of Mathematical Induction, (9) is valid for all $n\in \mathbb{N}^*$.

          Finally $(3+2\sqrt{2})^n\overset{(3)}{=}A_n+(\sqrt{2}B_n)\overset{(4)}{=}A_n+\left ( A_n-(3-2\sqrt{2})^n \right )=2A_n-(3-2\sqrt{2})^n=$

$$=(2A_n-1)+\left ( 1-(3-2\sqrt{2})^n \right ),$$

and since $0<1-(3-2\sqrt{2})^n<1$, for $n=1,\;2,\dots$, it results

$$x_n=[(3+2\sqrt{2})^n]=2A_n-1. \tag{10}$$

Then  $\frac{(x_n-1)(x_n+3)}{8}\overset{(10)}{=}\frac{(2A_n-2)(2A_n+2)}{8}=\frac{A_n^2-1}{2}\overset{(9)}{=}\frac{(2B_n^2+1)-1}{2}=B_n^2.$

QED $\blacksquare$

Problem E:17 120 - celebrating my 65th birthday

 

(On the 65th anniversary: may-08-2025)

         " E:17120.  Determine all prime numbers $n$ that have the property that the

           numbers $n^2+6$ and $n^2-6$ are prime .

(Emil Gheorghiu, Timișoara)"

ANSWER CiP

$$n=5$$

                    Solution CiP

                Let's make some attempts:

$n=2\;\;:\;n^2+6=10-this\; is \;NOT \;prime,\;\;n^2-6=-2;$

$n=3\;\;:\;n^2+6=15-this\; is \;NOT \;prime,\;\;n^2-6=3;$

$n=5\;\;:\;n^2+6=31,\;\;n^2-6=19 \;-BOTH \;are\; prime\;\;;$

$n=7\;\;:\;n^2+6=55-this\; is \;NOT \;prime,\;\;n^2-6=43;$

$n=11\;\;:\;n^2+5=127,\;\;n^2-6=115-this\; is \;NOT \;prime:$

$n=13\;\;:n^2+6=175-this\; is \;NOT \;prime,\;\;n^2-6=163;\dots$

     Moreover, it is observed that for $n\geqslant 7$, one of the sought numbers is a multiple of 5.

          Any natural number has one of the forms $5k,\; 5k\pm1,\;5k\pm2,\;\;k\in \mathbb{N}$. 

          If $n=5k$, being prime it can only be 5. The situation is convenient.

          If $n=5k\pm1$ then 

$n^2-6=(25k^2\pm10k+1)-6=25k^2\pm10k-5=5\cdot (5k^2\pm2k-1)$ 

so it is a multiple of 5 and greater than 5 for $n\geqslant11$. None of the numbers are convenient.

          If  $n=5k\pm2$ then 

$n^2+6=(25k^2\pm20k+4)+6=25k^2\pm20k+10=5 \cdot (5k^2\pm4k+2)$

so it is a multiple of 5 and greater than 5 for $n\geqslant 7$ . None of the numbers are convenient.

          So the only prime number $n$ with the desired property is 5.

$\blacksquare$

          Remark CiP It was an attempt, without much success, to consider the cases $n=6k\pm1,\;\;k\in \mathbb{N}$, the only forms that give prime numbers $\geqslant 5$ and cases $n=2$ and $n=3$ should be treated separately.

     We are left with only the conclusion: 

Always one of the four numbers $12k(3k\pm1),\;\;12k(3k\pm1)+2$ is divisible by 5.

<end Rem>

Crossed ladders problem revisited

     

Find the area of ​​the figure denoted by ?, knowing the areas of the other colored triangles X, Y, Z.

 ANSWER CiP

$$\textbf{S}_?=\frac{X\cdot Z \cdot(X+2Y+Z)}{Y^2-X\cdot Z}$$

                         Solution CiP

          It is just a takeover from other sources, but the calculations are mine.

          From the web-page Crossed ladders problem , we find a formula

 the formula links several areas together. In our notation

$$\frac{1}{\textbf{S}_{AEC}}+\frac{1}{\textbf{S}_{ADC}}=\frac{1}{\textbf{S}_{AFC}}+\frac{1}{\textbf{S}_{ABC}}.$$

          Substituting above with our data we have

$\frac{1}{X+Y}+\frac{1}{Y+Z}=\frac{1}{Y}+\frac{1}{\textbf{S}_?+X+Y+Z}$

so  $\frac{1}{\textbf{S}_?+X+Y+Z}=\frac{1}{X+Y}+\frac{1}{Y+Z}-\frac{1}{Y}=\frac{Y(Y+Z)+Y(X+Y)-(X+Y)(Y+Z)}{Y(X+Y)(Y+Z)}=\frac{Y^2-XZ}{Y(Y+X)(Y+Z)}.$

Hence $\textbf{S}_?=\frac{Y(Y+X)(Y+Z)}{Y^2-XZ}-X-Y-Z$ and we only get the answer by doing some calculations.

$\blacksquare$

A "brand" problem Marius BURTEA from Alexandria

          This is problem  E:17125, from GMB 2/2025, proposed for 6th grade.  In translation

          

           "Determine the prime numbers  $p\geqslant 5$ for which 156 is written as the sum of

            three natural numbers, directly proportional to 2,3 and $p$." 

(Marius BURTEA, Alexandria)

ANSWER CiP

$p=7\:\;:\;26+39+91=156;$

$p=47\;:\;6+9+141=156;$

$p=73\;:\;4+6+146=156:$

$p=151:\;2+3+151=156.$

                                 Solution CiP

          $\frac{a}{2}=\frac{b}{3}=\frac{c}{p}=\frac{a+b+c}{p+5}=\frac{156}{p+5},\;\;a,b,c \in \mathbb{N}$.

From the above we obtain

 $a=\frac{312}{p+5},\;\;b=\frac{468}{p+5},\;\;c=\frac{156p}{p+5}=\frac{156p+780-780}{p+5}=156-\frac{780}{p+5}.$

For the numbers $a,\;b,\;c$ above to be natural, $p+5$ must divide all the numbers 312, 468, 780. So $p+5$ must divide their greatest common divisor, that is

 $p+5 \mid gcd(312,468,780)=156 \tag{D}$.

     Given that $p\geqslant 5$ is a prime number, and that $p+5$ is an even number, among the numbers that verify condition (D), we choose

$p+5 \in \{12,\;26,\;52,\;78,\;156\}.$

Therefore $p\in \{7,\;21,\;47,\;73,\;151\}$ from which we eliminate the inconvenient value 21.

$\blacksquare$

A beautiful exponential equation

 

                     Topic 2. Find the real numbers $x$ for which

$3^x+3^{[x]}+3^{\{x\}}=4 \tag{E}$

                 ($[x]$ and $\{x\}$ represent the integer part and the fractional part of the real

                  number $x$, respectively)

ANSWER CiP

$$x_1=1-log_32,\;x_2=log_3{11}-2log_32-1$$

                             Solution CiP

              Because $x=[x]+\{x\}$ the equation is written equivalently

$3^{[x]} \cdot 3^{\{x\}}+3^{[x]}+3^{\{x\}}=4\;\;\;|+1\;\Leftrightarrow$

$\Leftrightarrow\;(3^{[x]}+1)\cdot (3^{\{x\}}+1)=5.$

          We obtain from the above  $3^{\{x\}}+1=\frac{5}{3^{[x]}+1}$  and because \frac $0\leqslant \{x\}<1$, therefore  $2\leqslant 3^{\{x\}}+1<4$, we must have

$$2\leqslant \frac{5}{3^{[x]}+1}<4.$$

The last inequalities imply

  $\frac{1}{4}<3^{[x]}\leqslant \frac{3}{2}$ 

and because  $[x]\in \{\dots,-2,\;-1,\;0,\;1,\;\dots\}$, it results $[x]=0$ or $[x]=-1$.

          If $[x]=0$, then $\{x\}=x$, and from (E) we obtain $3^x+1+3^x=4$ so $3^x=\frac{3}{2}$, hence $x=log_3\left ({\frac{3}{2}}\right )=log_33-log_32=1-log_32.$

If  $[x]=-1$, then $\{x\}=x+1$, and from (E) we obtain $3^x+\frac{1}{3}+3\cdot 3^x=4$ so $3^x=\frac{11}{12}$, hence $x=log_3 \left (\frac{11}{12} \right )=log_3{11}-log_3{(2^2\cdot 3)}=log_3{11}-2log_32-1$.

     Both values ​​found for $x$ verify equation (E).

$\blacksquare$

                         Official Solution

               It is shown that if  $x\geqslant 1$ or if  $x<-1$ then there are NO solutions.

It remains to study the cases  $x\in [0,1)$, when $[x]=0$, and  $x\in [-1,0)$, when  $[x]=-1$, and find, as with us, the answer.

Problem S:E25.67

         In tranlation

                              " S:E25.67. Knowing that $x+\frac{1}{x}=7$,  calculate $\frac{1}{x}-\frac{1}{x-7}.$"
 

ANSWER CiP

$$\frac{1}{x}-\frac{1}{x-7}=7$$

                            Solution CiP

               $x+\frac{1}{x}=7\;\Leftrightarrow\;x^2-7x+1=0 \tag{1}$

Now

$$\frac{1}{x}-\frac{1}{x-7}\underset{(1)}{=}7-x-\frac{1}{x-7}=7-\frac{x^2-7x+1}{x-7}\underset{(1)}{=}7-\frac{0}{x-7}=7$$

$\blacksquare$

           Remark CiP 

          The question is whether there are other expressions like this, $\frac{1}{x+m}+\frac{n}{px+q}$  which can be calculated from the given condition $x+\frac{1}{x}=7.$