A cute non-UFD ring

                You can quickly read about UFD here. (The author "Herstein", mentioned in the paragraph following Definition 4, seems to be I. N. HERSTEIN - Topics in Algebra - JOHN WILEY & SONS, 1975, Theorem at page 148)

               Let the ring $R=\mathbb{Z}_8[X]$. Units in $R$ are $\{\hat{1},\hat{3},\hat{5},\hat{7}\}$.

          The polynomial $X^2-\hat{1}$ admits two decompositions into irreducible factors:

$$X^2-\hat{1}\;=\;(X-\hat{1})(X+\hat{1})\;=\;(X-\hat{3})(X+\hat{3}). \tag{1}$$

The factor $X-\hat{1}$ is not associated with any of the factors $X\pm \hat{3}$. Neither does the other one $X+\hat{1}$. Indeed, $\hat{3}\cdot (X-\hat{1})=\hat{3}X-\hat{3}\neq X\pm \hat{3},\;\hat{5}\cdot (X-\hat{1})=-\hat{3}X+\hat{3}\neq X\pm\hat{3},\;$

$\hat{7}\cdot (X-\hat{1})=-X+\hat{1}\neq X\pm \hat{3}.$

          It can also be seen from (1) that the polynomial $X^2-\hat{1}$, although of degree two, has four roots in $\mathbb{Z}_8$.

               The example is taken from Michael ARTIN's book "ALGEBRA" (PRENTICE HALL, 1991) page 392, the example following Proposition (1.8).

QUO VADIS, Olimpiada Națională „Gazeta Matematică” ?

               In the past, it was written on the cover of GMB: "by solving problems from the math magazine, you also prepare for math competitions".

               The password to open the file is : ogeometrie . Just click on the image.

               Other information about the Mathematics Competitions in Romania can be found here.

               A collection of "Gazeta Matematica Seria B" can be found here.

MY ERRATA , MY FRIEND : the most harmless math mistakes

           I like to look for mistakes in other people's works. So, I don't have time to find my own mistakes.

           I was reading a wonderful book yesterday: 

My Numbers, My Friends (Popular Lectures on Number Theory) by the author  Paulo Ribenboim.

A first mistake, more of a haste than a typo, appears on page 10 (see marked text).

In the marked place, the correct text is :

$$...\;if\;p\;is\;a\;prime\;dividing\;both\;P,\;Q,\;then...$$

$\blacksquare$

Those who wish can read as much as they want by clicking on the image of the book cover.

GAZETA MATEMATICĂ Seria B N0 2/2024

 Click on the image to download.

For a wider collection click here. The password for opening encrypted files is    ogeometrie

SUPLIMENTUL cu EXERCIȚII al GMB N0 2/2024

 Click on the image to download.

For a wider collection click here. The password for opening encrypted files is    ogeometrie

It is not sure if I will be able to update ERATA with each issue of the magazine

           Nu este sigur daca voi reusi actualizarea ERATA la fiecare numar al revistei.

Тэнцүү нийлбэр бүхий хуваалтууд // Partitions with equal sums

           For which values of $n$ can the set $\{1,2,...,n\}$ be partitioned into three sets with the same sums of elements ?

       E.g $\{1,2,3,4,5\}=\{1,4\}\cup \{2,3\}\cup \{5\}$ ;

             $\{1,2,3,4,5,6\}=\{1,6\}\cup \{2,5\} \cup \{3,4\}.$

          The following is one of the old contest problems (22nd All Soviet Union Math Contest, 1988).

          Let $m,\; n,\; k$ be positive integers with $m \geqslant n$ and $1+2+\cdots +n=m\cdot k$. Prove that the numbers $1,\;2,\;\cdots,n$ can be divided into $k$ groups in such a way that the sum of the numbers in each group equals $m$.

          CiP example: $1+2+3+4+5+6+7+8+9=45$; 

          from $45=9\cdot 5$ the partition of $k=5$ groups is obtained, each with a sum of $m=9$

$$\{1,\;8\}\cup \{2,\;7\}\cup\{3,\;6\}\cup \{4,\;5\}\cup\{9\};$$

          from $45=15 \cdot 3$ the partition of $k=3$ goups is obtained, each with a sum of $m=15$.

$$\{1,\;2,\;3,\;9\} \cup \{4,\;5,\;6\} \cup \{7,\;8\}.$$

          CiP exampl$e^{bis}$: $1+2+3+4+5+6+7+8=36=12 \cdot 3$ and we have the partition of $k=3$ subsets, with the sum of  $m=12$

$$\{1,\;2,\;3,\;6\} \cup \{4,\;8\} \cup \{5,\;7\}.$$

 

          It would seem that the answer to the original question is $n=3\cdot p-1$ or $n=3\cdot p$. That is, in the case of $n=3\cdot p+1$, there is no such partition. (Indeed, if $n=3\cdot p+1$, then, because  $$1+2+ \cdots +n=\frac{n(n+1)}{2}=\frac{(3p+1)(3p+2)}{2}=9\cdot \frac{p(p+1)}{2}+1,$$ this sum should be a multiple of 3, which is not true.)

   If $n=3p$ then $1+2+ \cdots +n=\frac{3p(3p+1)}{2}=3 \cdot \frac{p(3p+1)}{2}=3\cdot \frac{p(p+1+2p)}{2}=3\cdot m$,

 and $p$ or $p+1$ is divisible by $2$ so $m$ is integer.

   If $n=3p-1$ then $\frac{n(n+1)}{2}=\frac{(3p-1) p}{2}=\frac {(2p+p-1)\cdot p}{2}=3 \cdot m$

with $m$ integer by the same arguments.

     So in these cases we fall over the general problem given to the 22nd All Soviet Union Math Contest, 1988.

          Another example in the case of $n=15$ is here : math.stackexchange.com/questions/31929/partition-of-equal-size-and-equal-sum

          The proof is given by induction, resulting in a way of constructing the partitions.

            The general problem has an object if $n \geqslant 3$.

          For $n=3$ we have only one partition $\{1,\;2\} \cup \{3\}.$

         For $n=4$ such partitions do not exsist.

         For $n=5,\;6$ are just the examples presented previously.

         For $n=7$ such partitions do not exist.

         For $n=8$, because $1+2+\cdots +8=36=9 \cdot 4=12 \cdot 3=18 \cdot 2$ we have in addition to the presented previously and the partitions

$$\{1,\;2,\;3,\;4,\;8\}\cup \{5,\;6,\;7\},\;\{1,\;3,\;6,\;8\}\cup \{2,\;4,\;5,\;7\}$$

$$\{1,\;4,\;5,\;8\}\cup \{2,\;3,\;6,\;7\}\;,\;\{1,\;4,\;6,\;7\}\cup \{2,\;3,\;5,\;8\}$$

$$\{1,\;2,\;4,\;5,\;6\}\cup \{3,\;7,\;8\}\;,\;\{1,\;2,\;3,\;5,\;7\} \cup \{4,\;6,\;8\}$$

$$\{1,\;2,\;7,\;8\}\cup \{3,\;4,\;5,\;6\}.$$

         For $n=9$, because $1+2+\cdots+9=45=9\cdot 5=15 \cdot 3$ are just the examples presented previously.