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JBMO 2007 problema 4

 
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Laurian Filip
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Joined: 25 Nov 2007
Posts: 473
Location: Bucuresti

PostPosted: Wed Apr 09, 2008 9:16 am    Post subject: JBMO 2007 problema 4 Reply with quote

Sa se arate ca daca p este un numar prim, atunci 7p+3^p-4 nu este patrat perfect.
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Ahiles
Euclid


Joined: 17 Apr 2008
Posts: 28

PostPosted: Thu Apr 17, 2008 7:12 pm    Post subject: Reply with quote

1) pentru p=2, avem 19 nu este patrat perfect.
2) p=3, avem 43 nu este patrat perfect.
3) p>3. Presupunem ca 7p+3^p-4=x^2.
Avem 3^p\equiv3 \pmod{p}, deci 7p+3^p-4\equiv -1 \pmod{p} sau x^2\equiv -1 \pmod{p} .
x^{p-1}\equiv 1 \pmod{p}
(x^2)^{\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}\equiv 1 \pmod{p} , deci \frac{p-1}{2}=2k sau p=4k+1.
7p\equiv 3 \pmod{4}
3^p\equiv 3 \pmod{4}
-4\equiv 0 \pmod{4}
De aici x^2\equiv 2 \pmod{4}, dar x^2 \equiv 0,1 \pmod{4}, contradictie.


Last edited by Ahiles on Fri Apr 18, 2008 6:54 pm; edited 3 times in total
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