A Divisibility Problem That Catches Us in the Morning

                It is Problem S:E24.359 from SGMB magazine (12/2024 at page 8), proposed for 8th grade, by a well-known author.

In translation:

              "Let $x$ and $y$ be nonzero natural numbers for which $13\mid (3x-2y)$. Show 

                 that the fraction

$\frac{31^{8(x+y)+1}}{x^2+y^2}$ is reducible."

This problem led me to make the Post here. 

ANSWER CiP

The given fraction can always be simplified by 13 

               Solution CiP

               Lemma 1  $13\mid (3x-2y) \Rightarrow 13\mid (5x+y)$ and $13\mid (x-5y).$

          Proof of  L1  We observe relationships:

$$2\cdot (5x+y)=13\cdot x-(3x-2y)\;\;;\;\;3\cdot (x-5y)=(3x-2y)-13 \cdot y.$$

From the hypothesis $13\mid (3x-2y)$ it then follows that

$$13 \mid 2\cdot (5x+y)\; and\; 13\mid 3\cdot (x-5y)$$

and, because $(13\;,2)=1=(13\;,3)$, it must $13\mid (5x+y) \;and\; 13 \mid (x-5y)$.

qed_L1 $\square$

          Remark CiP I found the property in the mentioned post. <end Rem>

               Lemma 2  $13\mid (3x-2y) \Rightarrow 13\mid (x^2+y^2).$

          Proof of L2  We have the equation, easy to verify by calculations

$$(5x+y)^2+(x-5y)^2=26 \cdot (x^2+y^2).\tag{1}$$

From the hypothesis $13\mid (3x-2y)$ we have, applying Lemma 1 :

 $13^2 \mid (5x+y)^2\;and\;13^2 \mid (x-5y)^2\;\;\overset{(1)}{\Rightarrow} 13^2\mid 26\cdot (x^2+y^2)\Rightarrow 13\mid 2\cdot(x^2+y^2)$

and, because $(13\;,2)=1$ it must
$$13\mid (x^2+y^2). \tag{2}$$

qed_L2 $\square$

          In the following equations the number $k$ has different values, which we are not interested in:

$31=13\cdot k+5\Rightarrow 31^2=31\cdot(13k+5)=13k+155=13k+12\cdot 13-1=13k-1\Rightarrow$

$\Rightarrow 31^3=31\cdot (13k-1)=13k-26-5=13k-5\Rightarrow 31^4=31\cdot(3k-5)=13k-12\cdot 13+1=13k+1.$

For $m=2(x+y)$ we obtain 

$31^{8(x+y)+1}-18=31^{4\cdot m+1}-18=31^{4\cdot m}\cdot 31-18=(31^4)^m\cdot 31-18=$

$=(13k+1)^m\cdot 31-18=(13k+1) \cdot 31-18=13k+31-18=13k+13=13\cdot k.$

So the numerator of the fraction in the statement is divisible by $13$. This together with (2) proves the answer.

$\blacksquare$

Other divisibility relations of 1-forms

               In post from April 9, 2020 I stated divisibility relations by 7 of some 1-forms. They are incomplete (even redundant), and should look like this:

$$7\mid\;x+2y\;\;\Leftrightarrow\;\;7\mid\;2x-3y\;\;\Leftrightarrow\;\;7\mid\;3x-y\; ;\tag{1}$$

$$7\mid\;x+3y\;\;\Leftrightarrow\;\;7\mid\;2x-y\;\;\Leftrightarrow\;\;7\mid\;3x+2y\;. \tag{2}$$

If in (1) we write $-y\;$ instead of $\;y\;$ we get

$$7\mid\;x-2y\;\;\Leftrightarrow\;\;7\mid\;2x+3y\;\;\Leftrightarrow\;\;7\mid\;3x+y \tag{1a}$$

that is, the second divisibility from the old post. So (1) and (1a) express the same thing. If in (2) we write $-y\;$ instead of $\;y\;$ we get

$$7\mid\;x-3y\;\;\Leftrightarrow\;\;7\mid\;2x+y\;\;\Leftrightarrow\;\;7\mid\;3x-2y. \tag{2a}$$

........................................................................................................................................

          Suggested by a problem elsewhere, we will state a set of divisibility equivalent to $13\mid 3x-2y.$ I got it for this:

$$13\mid 3x-2y\;\Leftrightarrow\;13\mid x-5y\;\Leftrightarrow\;13\mid 2x+3y\;\Leftrightarrow\;13 \mid 5x+y \tag{13a}$$

and by replacing $\;y$  with $\;-y$ :

$$13\mid 3x+2y\;\Leftrightarrow\;13\mid x+5y\;\Leftrightarrow\;13\mid 2x-3y\;\Leftrightarrow\;13\mid 5x-y. \tag{13a'}$$

The other relationships are:

$$13\mid  x-6y\;\Leftrightarrow\;13\mid 2x+y\;\Leftrightarrow\;13\mid 3x+5y\;\Leftrightarrow\;13\mid 5x-4y\;, \tag{13b}$$

$$13\mid x+6y\;\Leftrightarrow\:13\mid 2x-y\;\Leftrightarrow\;13\mid 3x-5y\;\Leftrightarrow\;13\mid 5x+4y\;, \tag{13b'}$$

$$13\mid x+2y\;\Leftrightarrow\;13\mid 4x-5y\;\Leftrightarrow\;13\mid 5x-3y\;\Leftrightarrow\;13\mid 6x-y\;, \tag{13c}$$

$$13\mid x-2y\;\Leftrightarrow\;13\mid 4x+5y\;\Leftrightarrow\;13\mid 5x+3y\;\Leftrightarrow\;13\mid 6x+y\;, \tag{13c'}$$

$13\mid x+3y\Leftrightarrow 13\mid 3x-4y\Leftrightarrow 13\mid 4x-y\Leftrightarrow 13\mid 5x+2y\Leftrightarrow 13\mid 6x+5y, \tag{13d}$

$13\mid x-3y\Leftrightarrow 13\mid 3x+4y\Leftrightarrow 13\mid4x+y\Leftrightarrow 13\mid 5x-2y\Leftrightarrow 13\mid 6x-5y, \tag{13d'}$

$13\mid x+4y\Leftrightarrow 13\mid 2x-5y\Leftrightarrow 13\mid 3x-y\Leftrightarrow 13\mid 4x+3y\Leftrightarrow 13\mid 5x-6y, \tag{13e}$

$13\mid x-4y\Leftrightarrow 13\mid 2x+5y\Leftrightarrow 13\mid 3x+y \Leftrightarrow 13 \mid 4x-3y\Leftrightarrow 13 \mid 5x+6y. \tag{13e'}$

          Let's demonstrate for example the equivalents from (13d).

$13\mid x+3y \Leftrightarrow \hat{1}\cdot x+\hat{3} \cdot y=\hat{0},$ where $\hat{m}$ mean the congruence class modulo 13  determined by $m$; their set is $\mathbb{Z}/13\mathbb{Z}$ (actually it is the field $GF(13)$. If we multiply last equation by $\hat{3},\;\hat{4},\;\hat{5},\;\hat{6}$ we obtain respectively

$$\hat{3}\cdot x +\hat{9}\cdot y=\hat{0} \Leftrightarrow \hat{3}\cdot x-\hat{4}\cdot y=\hat{0} \Leftrightarrow 13 \mid 3x-4y;$$

$$\hat{4} \cdot x+\hat{12} \cdot y=\hat{0} \Leftrightarrow \hat{4} \cdot x-\hat{1}\cdot y=\hat{0} \Leftrightarrow 13\mid 4x-y;$$

$$\hat{5}\cdot x+\hat{15} \cdot y=\hat{0} \Leftrightarrow \hat{5}\cdot x+\hat{2} \cdot y=\hat{0} \Leftrightarrow 13\mid 5x+2y;$$

$$\hat{6}\cdot x+\hat{18} \cdot y=\hat{0} \Leftrightarrow \hat{6} \cdot x+\hat{5} \cdot y=\hat{0} \Leftrightarrow 13\mid 6x+5y;$$

I used some elementary properties of $GF(13)$.

$\blacksquare$

A Problem with the Year of Birth Inserted into It

              About Problem S:E24.345 from the SGMB magazine 12/2024, proposed for the 7th grade. Authors Doina STOICA and Mircea_Mario STOICA (the latter born in 1961; we were classmates).

          In translation:

              "Let $\;a, \;b\;$ and $\;c\;$ be the lengths of the sides of triangle $ABC$. We know that

$$\sqrt{a^2-20a+2061}+\sqrt{b^2-16b+100}+\sqrt{c^2-12c+52}=10+\sqrt{1961}.$$

                Prove that triangle $ABC$ is right-angled."

ANSWER CiP

$a=10,\;\;b=8,\;\;c=6\;\;$ and   $a^2=b^2+c^2$

                    Solution CiP

               The given equation can also be written as

$$\sqrt{(a-10)^2+1961}+\sqrt{(b-8)^2+36}+\sqrt{(c-6)^2+16}=10+\sqrt{1961} \tag{1}$$

We have that LHS $\geqslant \sqrt{1961}+\sqrt{36}+\sqrt{16}=\sqrt{1961}+6+4=10+\sqrt{1961}.$

It means that in (1) we have equality, therefore $a=10,\;b=8,\;c=6$ and the triangle is the Ancient Egyptian one.

$\blacksquare$

E : 17 056 Everything is just zero

           It is Problem E:17056 of GMB 11/2024, page 651, authors Dragoș PETRICĂ and Cosmin MANEA from Pitești. In translation:

              "Determine the natural numbers $x,\;y,\;z,\;t\;$ and $u\;$ that simultaneously 

                 verify the relations  $x^2+y^2+z^2+t^2=7\cdot u^2$  and  $x\cdot t=y\cdot z$"

 ANSWER CiP

$$x=y=z=t=u=0$$

                       Solution CiP

               Taking into account the relationship

$$x\cdot t=y\cdot z \tag{1}$$

 the given equation is written successively equivalent to

$(x^2\pm 2 \cdot xt+t^2)+(y^2\mp 2 \cdot yz+z^2)=7\cdot u^2\;\;\Leftrightarrow$

$$\Leftrightarrow\;\;(x\pm t)^2+(y\mp z)^2=7\cdot u^2. \tag{2}$$

From (2) it is seen that the prime number $7$ divides a sum of two squares of integers. We will first prove the following

           Lemma  For  $A,\;B \in \mathbb{Z}$  we have   $7\;\mid\;A^2+B^2\;\;\Rightarrow\;\;7\mid A\;\;and\;\;7\mid B.$

            Proof of the Lemma 

            We have, after the remainders of dividing $A$ by $7$, one of the possibilities

$$(i)\;A=7 \cdot k,\;\;(ii)\;A=7\cdot k\pm 1,\;\;(iii)\;A=7 \cdot k \pm 2,\;\;(iv)\;A=7\cdot k \pm3,\;\;\; k\in \mathbb{Z}$$

Then, we have respectively $(i)\;A^2=(7k)^2=7\cdot(7k^2)=7\cdot K$

$(ii)\;A^2=(7k\pm 1)^2=49k^2\pm 14k+1=7\cdot K+1$

$(iii)\;A^2=(7k\pm 2)^2=49k^2\pm 28k+4=7\cdot K+4=7\cdot(K+1)-3$

$(iv)\;A^2=(7k\pm 3)^2=49k^2 \pm 42k+9=7\cdot (7k^2\pm 6k+1)+2$

where, in different cases, the number $K$ has different values. In other words, the natural remainders of dividing $A^2$ by $7$ can only be  $0,\;1,\;2\; or\; 4.$ Considering instead of the remainder $4$ the equivalent value $-3$, $(mod\; 7)$, we have one of the possibilities

$$A^2=7\cdot K,\;\;A^2=7 \cdot K+1,\;\;A^2=7\cdot K+2,\;\;A^2=7\cdot K-3,\;\;\;K\in\mathbb{N}.$$

     Obviously we have the same cases for $B^2$, and then, for $A^2+B^2$  we have the possibilities in the Table below, also calculated mod_7:

$$\begin{array}{|c|c|c|c|c|}\hline B^2/A^2&0&1&2&-3\\ \hline  0&0&1&2&-3\\ \hline  1&1&2&3&-2\\ \hline 2&2&3&-3&-1\\ \hline -3&-3&-2&-1&2\\ \hline\end{array}$$

            It can be seen from the Table above that $7\mid A^2+B^2$, that is $A^2+B^2=0\;mod\;7$ if and only if $A^2=0\;mod\;7\;and\;\;B^2=0\;mod\;7$, or $7\mid A^2\;\;and\;\;7\mid B^2$, and hence $7\mid A\;\;and\;\;7\mid B.$

$\square\;Lemma$

 

          Returning to (2), we have, according to Lemma

$$|x\pm t|=7\cdot v,\;\;|y\mp z|=7\cdot w,\;\;\;v,\;w \in \mathbb{N}\;;\;\Rightarrow$$

$\Rightarrow\;49(v^2+w^2)=7u^2\;\;\Leftrightarrow\;\;7(v^2+w^2)=u^2$, and hence $7\mid u$, so $u=7\dot u_1$ and

$$v^2+w^2=7\cdot u_1^2,\;\;\;u_1<u,\;\;u_1\in\mathbb{N}.\tag{3}$$

But in equation (3) we see that $7\mid (v^2+w^2)$ so, according Lemma again, $7\mid v$ and $7\mid w$. Then $v=7\cdot v_1,\;w=7\cdot w_1$ and $$49 \cdot v_1^2+49 \cdot w_1^2=7\cdot u_1^2\;\;\Leftrightarrow\;\;7(v_1^2+w_1^2)=u_1^2.$$

From here it follows $7\mid u_1$ so $u_1=7\cdot u_2$, and

$$v_1^2+w_1^2=7 \cdot u_2^2,\;\;\;u_2<u_1,\;\;u_2\in\mathbb{N}. \tag{4}$$

Comparing (3) and (4) we see that, continuing the process, we would obtain an infinitely decreasing sequence of natural numbers $u>u_1>u_2>\dots$, which is not possible. So $u=0.$

     But then $x^2+y^2+z^2+t^2=0$ and hence $x=y=z=t=0$. We got the answer.

 $\blacksquare$

          Remark CiP  Although $7$ is a sum of four squares $7=2^2+1^1+1^2+1^2$  the condition (1) is not met.

<end Rem>

Probably the HARDEST Geometry problem

                "Consider $n$ angles around a point, any two of which have distinct 

                 measures. Expressed in degrees, the measures of the $n$ angles are, 

                 any two, coprime natural numbers. Find the maximum value of $n$."

         Problem appeared in GMB 11/2024, page 648, proposed for 6th grade(E:17038), author Traian PREDA from Bucharest.

I don't have the ANSWER

I found examples with  ${\color{Green}{n=15}}$ 

and I would be happy if this was the maximum required.

          A few heuristic estimates will give us a first picture of the phenomenon.

         1) Because the angles around a point add up to $360^{\circ}$ and

$$1+2+3+\dots+26=351<360<378=1+2+3+\dots+26+27$$

we have the condition

$$3\leqslant n \leqslant 26 \tag{1}$$

          2) At most one of the angles has an even number of degrees as its measure.

          3) If we want as many angles as possible, then their sizes should be as small as possible.

So let's try prime numbers.

$1^{\circ}+2^{\circ}+3^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+41^{\circ}+$

$$+43^{\circ}+47^{\circ}+53^{\circ}=382^{\circ}$$

The result is $22^{\circ}$  more than it should be. We can eliminate $3^{\circ}+19^{\circ}$ or $5^{\circ}+17^{\circ}$ from the above sum and obtain

$$1^{\circ}+2^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+23^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+41^{\circ}+$$

$$+43^{\circ}+47^{\circ}+53^{\circ}=360^{\circ}$$

or

$$1^{\circ}+2^{\circ}+3^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+41^{\circ}+$$
$$+43^{\circ}+47^{\circ}+53^{\circ}=360^{\circ}.$$

Both writings have $n=15$ terms.

        By mistake, I wrote the terms (forgetting 47) but intentionally leaving out the even number 2

$$1^{\circ}+3^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+$$

$$+41^{\circ}+43^{\circ}+53^{\circ}=333^{\circ}$$

We would still have to add  $27^{\circ}$ to $360^{\circ}$. But then we would have to eliminate $3^{\circ}$ and instead of  $1^{\circ}+3^{\circ}$ we put  $4^{\circ}$. Therefore

$$4^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+$$

$$+27^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+41^{\circ}+43^{\circ}+53^{\circ}=360^{\circ}$$

So again a sum of  $n=15$  terms.

           As if with even more magic, I obtained another option. I see that

$$1^{\circ}+2^{\circ}+3^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+$$

$$+31^{\circ}+37^{\circ}+41^{\circ}+43^{\circ}+47^{\circ}=329^{\circ}$$

We would still need to add  $31^{\circ}$  to  $360^{\circ}$. We have the term $31^{\circ}$, but since $31+31=62$, close to 64 (!!) I will eliminate the term $31^{\circ}$ and replace it with $64^{\circ}$. We obtained a result with 2 more 360, so we will also eliminate the term $2^{\circ}$. Therefore

$$1^{\circ}+3^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+37^{\circ}+$$

$$+41^{\circ}+43^{\circ}+47^{\circ}+64^{\circ}=360^{\circ}$$

again a sum of  $n=15$  terms.

<last edit Ian, 23, 2025>

<will it continue ???>

Problem E : 17 052

             In GMB 11/2024 page 650,  author Mihaela BERINDEANU, Bucharest, proposed for 8th grade.

In translation:

"Solve in $\mathbb{Z}$ the equation $x^2+2y^2-4=2xy-2|x-y|.$"

ANSWER CiP

$$(x,y) \in \{(-2,-2),\;(-2,-1),\;(0,\pm 1),\;(2,1),\;(2,2)\}$$

                    Solution CiP

               Let us denote $z=x-y$. We have 

$$x=y+z \tag{1}$$

and the equation is written equivalently

  $$(y+z)^2+2y^2-4=2(y+z)y-2|z|\;\Leftrightarrow$$

$$\Leftrightarrow\;\;y^2+2yz+z^2+2y^2-4=2y^2+2yz-2|z|\;\Leftrightarrow\;y^2+z^2+2|z|-4=0\;\Leftrightarrow$$

$$\Leftrightarrow\;\;y^2+(|z|+1)^2=5. \tag{2}$$

In $\mathbb{Z}$ equation (2) is satisfied in the cases

$\begin{cases}y^2=1\\|z|+1=2; \end{cases} \tag{3}$

$\begin{cases}y^2=4\\|z|+1=1. \end{cases} \tag{4}$

Case (3) gives us $$\begin{cases}y=\pm1\\x-y=\pm1 \end{cases}$$

and we find the solutions $(2,1),\;(0,1),\;(0,-1),\;(-2,-1)$.

Case (4) give us $$\begin{cases}y=\pm 2\\x-y=0 \end{cases}$$

and we find the solutions $(2,2),\;(-2,-2)$. We got the answer.

$\blacksquare$

'Neața NEAȚĂ !! again... Problem S:E24.314

           Proposed for 8th grade by the same author; in SGMB 11/2024.

     In translation:

          "If $x$ and $y$ are nonzero integers such that $13(x^2+y^2)=(2x+3y)^2$, show

          that the fraction $\frac{3x+11y}{x^4+y^4+15x^2y^2}$ is reducible."

 

ANSWER CiP

$x=2 \cdot k,\;y=3 \cdot k,\;\;k \in \mathbb{Z}^*$ and 

the fraction always simplifies (at least) to $13$

                    Solution CiP

          The given condition about the numbers $x$ and $y$ is successively equivalent to

$13 \cdot x^2+13 \cdot y^2=4 \cdot x^2+12 \cdot xy+9 \cdot y^2\;\;\Leftrightarrow\;\;9x^2-12xy+4y^2=0\;\;\Leftrightarrow$

$$\Leftrightarrow\;\;(3x-2y)^2=0\;\;\Leftrightarrow\;\;3 \cdot x=2 \cdot y\;.$$

We see from the last equation that $2 \mid 3\cdot x$, but(EUCLD's Lemma) $2$ and $3$ being relatively prime, the result is that $2 \mid x$, so $x=2\cdot k$  fore some  $0 \neq k\in \mathbb{Z}$ and then 

$$3\cdot x=2\cdot y \;\Rightarrow\;\;3 \cdot 2k=2 \cdot y\;\;\Rightarrow y=3k\;.$$

          In this case the given fraction is

$$\frac{3\cdot 2k+11 \cdot 3k}{16k^4+81k^4+15 \cdot 4k^2\cdot 9k^2}=\frac{39 \cdot k}{637 \cdot k^4}=\frac{13\cdot 3k}{13 \cdot 49k^4}$$

and we get the answer.

$\blacksquare$

Vi er ikke alle så forskjellige, heldigvis... Problem E:17 053

          From GMB 11/2024 on page 650, proposed for 8th grade, author Constantin NICOLAU, Curtea_de_Argeș. In translation:

          "Let $x,\;y,\;z\;$ be nonzero real numbers, such that $x+y+z=24\;$ and

             $xy+yz+zx=192.\;$  Calculate $\frac{x^4}{y}+\frac{y^4}{z}+\frac{z^4}{x}.$"

ANSWER CiP

1 536     $(x=y=z=8)$

                    Solution CiP

          From $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$ we obtain

$$24^2=x^2+y^2+z^2+2 \cdot 192$$

$\Rightarrow\;\; x^2+y^2+z^2=576-384=192$ so
$$x^2+y^2+z^2=xy+yz+zx. \tag{1}$$

But (1)$\;\Leftrightarrow\;\;2\cdot x^2+2\cdot y^2+2\cdot z^2-2xy-2yz-2zx=0\;\Leftrightarrow$

$\Leftrightarrow\;\;(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)=0\;\Leftrightarrow$

$\Leftrightarrow\;\;(x-y)^2+(y-z)^2+(z-x)^2=0\;$ and this is possible in real numbers (being a sum of non-negative terms) only if $x-y=0=y-z=z-x.$ Therefore $x=y=z$; moreover $x=y=z=\frac{x+y+z}{3}=\frac{24}{3}=8.$

     We then obtain $\frac{x^4}{y}+\frac{y^4}{z}+\frac{z^4}{x}=3\cdot x^3=3\cdot 8^3=1536.$

$\blacksquare$

A SYCOPHANTIC EQUATION

      ... sorry, Diophantine !

          This is Exercise S:E24.305 from a magazine dear to us. The authors are Răzvan LUPU and Roxana VASILE from CRAIOVA. The exercise is proposed for 7th grade. In translation:

          "Determine the integers $x$ and $y$ that verify the relationship

$$x^2-xy+2x-3y=6."$$

ANSWER CiP

$$(x,y)\in \{(0,-2),\;(-2,-6),\;(-4,-2),\;(-6,-6)\}$$

Solution CiP

               We write the equation successively
$$x^2+3x-x-3+3-xy-3y=6\;\;\Leftrightarrow\;\;x(x+3)-(x+3)+3-y(x+3)=6\;\;\Leftrightarrow$$

$$\Leftrightarrow \;\;(x+3)(x-1-y)=3$$

Because $(x,y)\in \mathbb{Z}\times \mathbb{Z}$, we have the cases indicated on the columns in the table:

$$\begin{array}{c||c|c|c|c}x+3&1&-1&3&-3\\ \hline x-1-y&3&-3&1&-1\\ \hline \hline x&-2&-4&0&-6 \\ \hline y&-6&-2&-2&-6 \end{array}$$

We got the answer.

$\blacksquare$

     Remark CiP We guessed the decomposition by following the procedure of completing squares.

$$x^2-xy+2x-3y=x^2-x(y-2)-3y=x^2-2\cdot x\cdot \frac{y-2}{2}-3y=$$

$$=\left (x-\frac{y-2}{2}\right )^2-\underline{\left ( \frac{y-2}{2} \right )^2}-3y=\left (x-\frac{1}{2}y+1 \right)^2-\frac{1}{4}y^2-2y-1=$$

$$=\left(x-\frac{1}{2}y+1 \right )^2-\frac{1}{4}\cdot (y^2+8y+16)+\underline{4}-1=\left (x-\frac{1}{2}y+1 \right )^2-\frac{1}{4}(y+4)^2+3=$$

$$=\left [ \left (x-\frac{1}{2}y+1\right )-\frac{1}{2}(y+4) \right ]\cdot \left [ \left (x-\frac{1}{2}y+1 \right )+\frac{1}{2}(y+4) \right ]+3=$$

$$=(x-y-1) \cdot (x+3)+3...$$

<end Rem>

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

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

'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.

$\blacksquare$

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.

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