Home/ Intrebari/Q 80337
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

coco2000
Pe: 2013-02-06T16:55:58+02:00 In: In:

problemă olimpiada patrate perfecte

fie a si b două numere întregi astfel încât
4a^2-3b^2+1=7b-9a.
Să se arate că numerele 3a+3b+7 si 4a+4b+9 sunt pătrate perfecte.

atasez si incercarea mea de rezolvare dar nu iese !

Attached files

new_microsoft_office_word_document_205.zip (76.2 KB) 

Similare

2 raspunsuri

  1. bedrix
    expert (VI)
  2. Minecraft
    maestru (V)

        \[ \begin{array}{l} G.M.{\rm{ }}10;{\rm{ }}E] = (b - a)(3b + 3a + 7) \\ 4a^2 - 4b^2 + b^2 + 1 = 9b - 2b - 9a \Leftrightarrow \underbrace {b^2 + 1^2 + 2b}_{} = 4b^2 - 4a^2 + 9b - 9a = 4(b^2 - a^2 ) + 9(b - a) = 4(b + a)(b - a) + 9(b - a) = (b - a)\left[ {4(b + a) + 9} \right] = (b - a)(4b + 4a + 9) \\ \end{array} \right. \\ Daca{\rm{ }}4a^2 - 3b^2 + 1 = 7b - 9a \Leftrightarrow \left\{ \begin{array}{l} \underbrace {(a + 1)^2 }_{p.p.} = \underbrace {(b - a)(3a + 3b + 7)}_{p.p.} \\ \underbrace {(b + 1)^2 }_{p.p.} = \underbrace {(b - a)(4a + 4b + 9)}_{p.p.} \\ \end{array} \right.{\rm{ }}din{\rm{ }}egalitatea{\rm{ }}membrilor{\rm{ }}rezulta{\rm{ }}ca{\rm{ }}numerele{\rm{ }}\underbrace {(b - a)(3a + 3b + 7)}_{p.p.}{\rm{ }}si{\rm{ }}\underbrace {(b - a)(4a + 4b + 9)}_{p.p.}{\rm{ }}sunt{\rm{ }}patrate{\rm{ }}perfecte. \\ Dar{\rm{ }}daca{\rm{ }}ni{\rm{ }}se{\rm{ }}cere{\rm{ }}sa{\rm{ }}aratam{\rm{ }}ca{\rm{ }}(3a + 3b + 7){\rm{ }}si{\rm{ }}(4a + 4b + 9){\rm{ }}sunt{\rm{ }}patrate{\rm{ }}perfecte,{\rm{ }}atunci{\rm{ }}si{\rm{ }}numarul{\rm{ }}(b - a){\rm{ }}ar{\rm{ }}trebui{\rm{ }}sa{\rm{ }}fie{\rm{ }}p.p.{\rm{ }}de{\rm{ }}forma{\rm{ }}\left\{ \begin{array}{l} (a + 1)^2 = (b - a)^2 *m^2 {\rm{ }} = \left[ {(b - a)m} \right]^2 \\ (b + 1)^2 = (b - a)^2 *n^2 {\rm{ }} = \left[ {(b - a)n} \right]^2 \\ \end{array} \right. \\ Din\left\{ \begin{array}{l} (3a + 3b + 7) = 3\underbrace {(a + b + 2)}_k + 1 = 3k + 1 \\ (4a + 4b + 9) = 4\underbrace {(a + b + 2)}_k + 1 = 4k + 1 \\ \end{array} \right.{\rm{ }} \Rightarrow \left\{ \begin{array}{l} (b - a)(3k + 1) = p.p. \\ (b - a)(4k + 1) = p.p. \\ \end{array} \right.{\rm{ }}unde{\rm{ }}k,a,b \in Z*{\rm{ }} \\ Sa{\rm{ }}pre\sup unem{\rm{ }}ca{\rm{ }}(b - a){\rm{ }}nu{\rm{ }}e{\rm{ }}p.p.{\rm{ }}si{\rm{ }}il{\rm{ }}scriem{\rm{ }}ca{\rm{ }}produsul{\rm{ }}d{\mathop{\rm int}} re{\rm{ }}un{\rm{ }}patrat{\rm{ }}perfect{\rm{ }}si{\rm{ }}un{\rm{ }}numar{\rm{ }}{\mathop{\rm int}} reg,{\rm{ }}adica{\rm{ }}(b - a) = xy^2 ,{\rm{ }}unde{\rm{ }}x,y \in Z*{\rm{ }}si{\rm{ }}x \ne 1 \\ \underline {Observatie} :{\rm{ }}Pt.{\rm{ }}x = 1 \Rightarrow 1*y^2 = y^2 {\rm{ }}si{\rm{ }}de{\rm{ }}aici{\rm{ }}conditia{\rm{ }}x \ne 1. \\ Din\left\{ \begin{array}{l} (b - a)(3k + 1) = p.p. \Rightarrow xy^2 *(3k + 1) = p.p. \\ (b - a)(4k + 1) = p.p. \Rightarrow xy^2 *(4k + 1) = p.p. \\ \end{array} \right. \\ \left. \begin{array}{l} Daca{\rm{ }}(b - a) = xy^2 {\rm{ }}atunci{\rm{ }}(3k + 1) = xm^2 ,{\rm{ }}cu{\rm{ }}x \ne 1{\rm{ }} \Rightarrow (b - a)(3k + 1) = xy^2 *xm^2 = x^2 (ym)^2 {\rm{ }}unde{\rm{ }}x,y,m \in Z*{\rm{ }} \\ Daca{\rm{ }}(b - a) = xy^2 {\rm{ }}atunci{\rm{ }}(4k + 1) = xn^2 ,{\rm{ }}cu{\rm{ }}x \ne 1{\rm{ }} \Rightarrow (b - a)(4k + 1) = xy^2 *xn^2 = x^2 (yn)^2 unde{\rm{ }}x,y,n \in Z* \\ \end{array} \right\} \Rightarrow Pentru{\rm{ }}(b - a) \ne p.p{\rm{ }} \Rightarrow \left\{ \begin{array}{l} (3k + 1) = x*m^2 \\ (4k + 1) = x*n^2 \\ \end{array} \right.{\rm{ }} \\ \left. \begin{array}{l} Fie{\rm{ }}\underline d {\rm{ }}c.m.m.d.c.{\rm{ }}pt{\rm{ }}3k + 1{\rm{ }}si{\rm{ }}4k + 1{\rm{ }} \Rightarrow \left\{ \begin{array}{l} 3k + 1 = d*m^2 \\ 4k + 1 = d*n^2 \\ \end{array} \right.{\rm{ }}unde{\rm{ }}m^2 ,n^2 {\rm{ }}apartin{\rm{ }}lui{\rm{ }}Z*{\rm{ }}si{\rm{ }}\left( {m^2 ;n^2 } \right) = 1 \Rightarrow (m;n) = 1 \\ . \\ Daca\left\{ \begin{array}{l} d\left| {3k + 1} \right. \\ d\left| {4k + 1} \right. \\ \end{array} \right.{\rm{ }}atunci\left\{ \begin{array}{l} d\left| {4(3k + 1)} \right. \\ d\left| {3(4k + 1)} \right. \\ \end{array} \right.rezulta{\rm{ }}ca{\rm{ }}d\left| {(12k + 4)} \right. - (12k + 3) = 12k + 4 - 12k - 3 = 1{\rm{ }}rezulta{\rm{ }}ca{\rm{ }}d = \pm 1 \\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l} (3k + 1) = \pm 1*m^2 \\ (4k + 1) = \pm 1*n^2 \\ \end{array} \right.sunt{\rm{ }}prime{\rm{ }}{\mathop{\rm int}} re{\rm{ }}ele \\ Cum{\rm{ }}conditia{\rm{ }}era{\rm{ }}ca{\rm{ }}x \ne 1{\rm{ }}rezulta{\rm{ }}ca{\rm{ }}d = x = - 1{\rm{ }} \Rightarrow \left\{ \begin{array}{l} (b - a)(3k + 1) = x^2 (ym)^2 = \left( { - 1} \right)^2 *(ym)^2 = y^2 *m^2 \\ (b - a)(4k + 1) = x^2 (yn)^2 = \left( { - 1} \right)^2 *(yn)^2 = y^2 *n^2 \\ \end{array} \right. \\ {\mathop{\rm Re}\nolimits} zulta{\rm{ }}ca{\rm{ }}egalitatea{\rm{ }}\left\{ \begin{array}{l} \underbrace {(a + 1)^2 }_{p.p.} = \underbrace {(b - a)(3a + 3b + 7)}_{p.p.} \\ \underbrace {(b + 1)^2 }_{p.p.} = \underbrace {(b - a)(4a + 4b + 9)}_{p.p.} \\ \end{array} \right.{\rm{ }}este{\rm{ }}ade{\mathop{\rm var}} ata{\rm{ }}doar{\rm{ }}pt{\rm{ }}(b - a) = y^2 = p.p. \\ Iar{\rm{ }}daca{\rm{ }}(b - a) = p.p.{\rm{ }}atunci{\rm{ }}si{\rm{ }}numerele{\rm{ }}3a + 3b + 7{\rm{ }}si{\rm{ }}4a + 4b + 9{\rm{ }}sunt{\rm{ }}patrate{\rm{ }}perfecte,{\rm{ }}Q.E.D.(quod{\rm{ }}erat{\rm{ }}demonstrandum). \\ \end{array} \]

Raspunde

Raspunde