Am făcut o călătorie la Mediaș, orașul copilăriei soției, pentru niște mici cumpărături. Vreme așa și șa, se prognoza cam ploaie dar am scăpat...
...și mâncatul de rigoare...
.
.
Pe masă am ...nelipsita carte...
azi un Autor inedit :
Bogdan SUCEAVĂ.
Altfel, lumea este în mișcare...
Soția, mergând spre ATLASSIB, de două ori pe săptămână, marțea și vinerea
sau
la M/edicul de/ F/amilie/
(numai eu stiu care este ea, urcand in Autobuz)
Iar eu
S T A U
dar am un SUFLET DRAG pentru care merita sa traiesc...(care este pe Locul 2, dupa sotie...)
și îmi umplu timpul cu tot felul de Probleme...
"S : E26.250. Let $a,\;b,\;c\;$ be positive real numbers such that
$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{1}{a+b+c+2}$.
Prove that $\frac{\sqrt{abc+1}}{a+b+c+1}$ is a natural number.
{ Autor : } Ion NEAȚĂ, Slatina "
ANSWER CiP
$\frac{\sqrt{abc+1}}{a+b+c+1}=1$
Solution CiP
The given equation is written out step by step
$\frac{a+b+c}{abc}=\frac{1}{a+b+c+2}\;\Rightarrow\;abc=(a+b+c)(a+b+c+2)\;\Rightarrow$
$\Rightarrow\;abc+1=(a+b+c)^2+2(a+b+c)+1\;\Rightarrow\;abc+1=(a+b+c+1)^2\Rightarrow$
$\underset{a,\;b,\;c>0}{\Rightarrow}\;\sqrt{abc+1}=a+b+c+1\;\;\Rightarrow\;\;\frac{\sqrt{abc+1}}{a+b+c+1}=1\in\mathbb{N}$
$\blacksquare$
"S : L26.245. Let $a,\;b,\;c\;\in\mathbb{R}^*\;$ with $a+b+c=0$. Show that
$\frac{1}{-a^2+b^2+c^2}+\frac{1}{a^2-b^2+c^2}+\frac{1}{a^2+b^2-c^2}=0.$
{No Author}"
Solution CiP
The denominators are, under the given conditions, different from zero. Indeed,
$-a^2+b^2+c^2=-(-b-c)^2+b^2+c^2=-2bc\neq 0$
and the analogous expressions. Then
$\frac{1}{-a^2+b^2+c^2}+\frac{1}{a^2-b^2+c^2}+\frac{1}{a^2+b^2-c^2}=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}=$
$=\frac{1}{2}\cdot \left ( \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab} \right )=-\frac{1}{2}\cdot \frac{a+b+c}{abc}=-\frac{1}{2}\cdot \frac{0}{abc}=0.$
$\blacksquare\;\;\blacksquare$
"S: L26.254. Prove that the equation
$(x-1)(y-1)=\sqrt[3]{z}(x+y-\sqrt[3]{z})$
has infinitely many solutions in the set of natural numbers.
{Author : } Traian TĂMÂIAN, Carei"
ANSWER CiP
$x=y=2k^2-2k+1\;\;,\;z=8k^6\;or\;z=8(k-1)^6\;\;\;,\;k\in\mathbb{N}$
For example, some particular solutions are
$(1,\;1,\;8)\;,\;(5,\;5,\;8)\;,\;(5,\;5,\;2^9)\;,\;(13,\;13,\;2^9)\;,\;(13,\;13,\;8\cdot 3^6)$
Solution CiP
Let $z=u^3$ , for which the equation is written as
$(x-1)(y-1)=u(x+y-u)\;\Leftrightarrow$
$\Leftrightarrow\;\;u^2-u(x+y)+(x-1)(y-1)=0 \tag{1}$
The discriminant of the equation (1) with the unknown $u$ is
$\Delta_u=(x+y)^2-4(x-1)(y-1)=(x^2+2xy+y^2)-4(xy-x-y+1)=$
$=(x-y)^2+4(x+y-1)$
Let's try to find solutions where $x=y$. Then $\Delta_u=4(2x-1)$ , and for this to be a perfect square, we impose the condition
$2x-1=(2k-1)^2\;\;,\;k\in\mathbb{N} \tag{2}$
from which we obtain
$x=y=2k^2-2k+1 \tag{3}$
By virtue of (2) the solutions of the equation (1) will be
$u_{1,2}=\frac{x+y \pm2(2k-1)}{2}=x\pm(2k-1)\underset{(3)}{=}2k^2-2k+1\pm(2k-1)$
that is $u_1=2k^2\;\;,\;\;u_2=2k^2-2k+2=2(k-1)^2.$ from which it follows that
$z_1=8k^6\;\;,\;\;z_2=8(k-1)^6 \tag{4}$
Both values (4) , together with (3) , satisfy the equation :
$(i)\;\;(x-1)(y-1)=(2k^2-2k)^2=4k^2(k-1)^2$
$\sqrt[3]{z}(x+y-\sqrt[3]{z})=2k^2(4k^2-4k+2-2k^2)=4k^2(k^2-2k+1)=4k^2(k-1)^2$
$(ii)\;\;\sqrt[3]{z}(x+y-\sqrt[3]{z})=2(k-1)^2[4k^2-4k+2-2(k-1)^2]=$
$=2(k-1)^2(4k^2-4k+2-2k^2+4k-2)=2(k-1)^2\cdot 2k^2$
$\blacksquare\;\;\blacksquare\;\;\blacksquare$
