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>