PROBLEM OC531 - CRUX MATHEMATICORUM, V47n5

 Pag 238 - En

Pag 239 - Fr

ANSWER CiP

If $7 \mid k$ then solutions are $(k,0),\;(k,k),\;(\frac{3k}{7},-\frac{k}{7})$,

and, if any, the pairs $(x_1,y_1),\;(k-x_1+y_1,y_1).$

If $7 \nmid k$ then the solution $(\frac{3k}{7},-\frac{k}{7})$(although rational),

 is not made up of integers; the others remain valid.

Examples

 If $k=-4$, the equation has solutions

$(-4,0),\;(-4,-4),\;(-1,-3),\;(-6,-3).$

If $k=7$, the equation has solutions

$(7,0),\;(7,7),\;(3,-1),\;(3,6),\;(10,6).$ 

 

 

Solution CiP

          There are two pairs $(x,y)$ that obviously check the equation: $(k,0)$ and $(k,k)$.

          The given equation is equivalent to

$$x^2-xy+2y^2-kx-ky=0\; ;\; x+y \neq 0. \tag{1} \label{eq1}$$

           Let $(x_1,y_1)$ be a solution of equation (1), with the condition $x_1+y_1 \neq 0$. This means that the equation with unknow $x$

$$x^2-(k+y_1)x+2y_1^2-ky_1=0 \tag{2}$$

has the solution $x=x_1$. Being of the second degree for  $x$, she has another solution $x_2$ which, according to Viete's Relations, she checks $x_1+x_2=k+y_1$, so $x_2=k-x_1+y_1.$

     For the pair $(x_2,y_1)$ to fit the given problem it must be $x_2+y_1 \neq 0$. But $x_2+y_1=0 \; \Leftrightarrow \;k-x_1+2y_1=0 \; \Leftrightarrow \; x_1=k+2y_1$, and writing that $k+2y_1$ check equation (2) we immediately get $y_1=0$, then $x_1=k$.

          So for any solution $(x_1,y_1) \neq (k,0)$ of the equation (1) we get another solution by the transformation

$$(x_1,y_1)\; \mapsto \; (k-x_1+y_1,y_1). \tag{3}$$

       We have $(k,k) \overset{(3)}{\mapsto} (k-k+k,k)=(k,k)$, so by transforming (3) we do not get a NEW solution. Let's look for all the solutions that remain invariant through this transformation, so $k-x_1+y_1=x_1$, whence $y_1=2x_1-k$. If we write that $(x_1,2x_1-k)$ check equation (1)

$$x_1^2-x_1(2x_1-k)+2(2x_1-k)^2-kx_1-k(2x_1-k)=0$$

$\Leftrightarrow \;7x_1^2-10kx_1+3k^2=0$ we get $x_1=k$ or $x_1=\frac{3k}{7}$. In the secon case $y_1=-\frac{k}{7}$. So the only invariant pairs by transformation (3) are $(k,k)$ and $(\frac{3k}{7}, -\frac{k}{7})$.

          So, if $k \;\vdots \; 7$ (means "k is divisible by 7"), then the equation (1) has three solutions $(k,0),\; (k,k),\;(\frac{3k}{7},-\frac{k}{7})$ to which, if there are others $(x_1,y_1)$, an even number of solutions $(x_1,y_1)$ and $(k-x_1+y_1,y_1)$ is added. In total an odd number.

     Otherwise ($7 \nmid k$) the solution $(\frac{3k}{7},-\frac{k}{7})$ does not consits of integers, and the number of solutions is even.

$\blacksquare$

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Tracking Number: 12749
Received:        Wed Jul  7 08:52:45 2021

From:  Petre Ciobanu
       Scoala Gimnaziala "Samuil Micu" SADU
       Sibiu, Romania
Email: ptr.ciobanu@gmail.com

Type:  Solve an Olympiad Corner Problem
       (problem OC531)

Files:
  LATEX_source_for_PROBLEM_OC_531.doc
  Problem_OC_531.png


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Crux Mathematicorum <crux@cms.math.ca>	11:53 (acum 7 ore)
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