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JBMO 2008 problema 3

 
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Omer Cerrahoglu
Euclid


Joined: 17 Mar 2008
Posts: 34

PostPosted: Thu Jul 17, 2008 2:27 pm    Post subject: JBMO 2008 problema 3 Reply with quote

Determinati toate numerele prime p, q si r pentru care  \frac{p}{q}-\frac{4}{r+1}=1.
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Theodor Munteanu
Thales


Joined: 06 May 2008
Posts: 126
Location: Sighetu Marmatiei/Bucharest

PostPosted: Fri Jul 18, 2008 2:36 pm    Post subject: Reply with quote

Avem (p-q)(r+1)=4q;
daca r=2 rezulta p=7 si q=3;
daca r>2 rezulta p-q=2k si r+1=2k' deci kk'=q;
dar q e prim deci:
1)k=q;k'=1 si p-q=2q deci p=3q dar q>2 deci p>3 si nu e prim;
2)k=1 deci p-q=2 deci (q+2)/q-4/(r+1)=1 sau r=2q-1.
Astfel avem tripletele(q+2,q,2q-1) si (7,3,2).
Nu stiu de ce dar latexul nu merge asa ca am scris "de mana".[/code][/quote][/tex]
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Last edited by Theodor Munteanu on Fri Jul 18, 2008 7:18 pm; edited 3 times in total
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Marius Mainea
Gauss


Joined: 26 May 2008
Posts: 1099
Location: Gaesti (Dambovita)

PostPosted: Fri Jul 18, 2008 3:38 pm    Post subject: Reply with quote

Theodor Munteanu wrote:
{\rm 1)k = q si k' = 1;p - q = 2q} \Rightarrow {\rm p = 3q iar p > 3 deci p nu e prim;} \\
{\rm 2)k = 1 si k' = q;p - q = 2 si }\frac{{{\rm q + 2}}}{{\rm q}} - \frac{4}{{r + 1}} = 1 \Rightarrow r + 1 = 2q{\rm deci tripletele} \\
{\rm (p,q,r) = (q + 2,q,2q - 1)}{\rm .si (7,3,2)} \\
\end{array}
\]' title=' \[ \begin{array}{l} Avem{\rm p(r + 1) - 4q = q(r + 1) sau (p - q)(r + 1) = 4q,p} \ne q; \\ {\rm daca r = 2} \Rightarrow {\rm q = 3} \Rightarrow {\rm p = 7;} \\ {\rm r > 2} \Rightarrow {\rm p - q = 2k deci p,q > 2}{\rm .} \\ {\rm 2k*2k' = 4q} \Rightarrow {\rm kk' = q(r + 1 = 2k') deci :} \\ {\rm 1)k = q si k' = 1;p - q = 2q} \Rightarrow {\rm p = 3q iar p > 3 deci p nu e prim;} \\ {\rm 2)k = 1 si k' = q;p - q = 2 si }\frac{{{\rm q + 2}}}{{\rm q}} - \frac{4}{{r + 1}} = 1 \Rightarrow r + 1 = 2q{\rm deci tripletele} \\ {\rm (p,q,r) = (q + 2,q,2q - 1)}{\rm .si (7,3,2)} \\ \end{array} \]' alt=' \[ \begin{array}{l} Avem{\rm p(r + 1) - 4q = q(r + 1) sau (p - q)(r + 1) = 4q,p} \ne q; \\ {\rm daca r = 2} \Rightarrow {\rm q = 3} \Rightarrow {\rm p = 7;} \\ {\rm r > 2} \Rightarrow {\rm p - q = 2k deci p,q > 2}{\rm .} \\ {\rm 2k*2k' = 4q} \Rightarrow {\rm kk' = q(r + 1 = 2k') deci :} \\ {\rm 1)k = q si k' = 1;p - q = 2q} \Rightarrow {\rm p = 3q iar p > 3 deci p nu e prim;} \\ {\rm 2)k = 1 si k' = q;p - q = 2 si }\frac{{{\rm q + 2}}}{{\rm q}} - \frac{4}{{r + 1}} = 1 \Rightarrow r + 1 = 2q{\rm deci tripletele} \\ {\rm (p,q,r) = (q + 2,q,2q - 1)}{\rm .si (7,3,2)} \\ \end{array} \]' align=absmiddle>


Poti sa scrii putin mai clar?
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Omer Cerrahoglu
Euclid


Joined: 17 Mar 2008
Posts: 34

PostPosted: Sat Jul 19, 2008 4:13 pm    Post subject: Reply with quote

Theodor Munteanu wrote:
Avem (p-q)(r+1)=4q;
daca r=2 rezulta p=7 si q=3;
daca r>2 rezulta p-q=2k si r+1=2k' deci kk'=q;
dar q e prim deci:
1)k=q;k'=1 si p-q=2q deci p=3q dar q>2 deci p>3 si nu e prim;
2)k=1 deci p-q=2 deci (q+2)/q-4/(r+1)=1 sau r=2q-1.
Astfel avem tripletele(q+2,q,2q-1) si (7,3,2).

Ai 2 greseli in solutie:
1)Din r>2 nu rezulta  p-q=2k(qpoate lua valoarea 2)
2)Tripletul (q+2,q,2q-1) reaprezinta o singura solutie in numere prime(gandestete de ce Wink )
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Sorin
Euclid


Joined: 19 Jul 2008
Posts: 13
Location: Timisoara

PostPosted: Fri Aug 08, 2008 2:06 pm    Post subject: Reply with quote

pai da, de la (p-q)(r+1)=4q am luat si eu pe cazuri
(1) p-q=1 si r+1=4q \Leftrightarrow q=2;p=3;r=7
(2) p-q=2qsi r+1=2 \Leftrightarrow r=1,nu convine
(3) p-q=q si r+1=4 \Leftrightarrow p=2q \Leftrightarrow q=1,nu convine
(4) p-q=4 si r+1=q \Leftrightarrow q=p-4=r+1 \Leftrightarrow r=p-5 \Leftrightarrow p=7;q=3;r=2
(5) Cazul asta intr-adevar da cele mai multe batai de cap : p-q=2 si r+1=2q \Leftrightarrow p=q+2 si r=2q-1 .
Daca q e de forma 3k obtinem o singura solutie pentru k=1 : q=3,p=r=5
Daca q e de forma 3k+1, cu k \geq 2 p va fi de forma 3k+3=3(k+1) \vdots 3
Daca q e de forma 3k+2 r va fi de forma 6k+3=3(k+2) \vdots 3

Rezumand cele 5 cazuri obtinem (q,p,r)\in {(2,3,7);(3,7,2);(3,5,5)}
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