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.

<end Rem>