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

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

$\blacksquare$