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JBTST III 2008, Problema 3

Posted: Mon Jun 16, 2008 11:47 pm
by Laurian Filip
Sa se determine numerele prime \( p \) si \( q \) ce verifica relatia

\( 2p^q-q^p=7. \)

Posted: Wed Jun 18, 2008 10:04 pm
by Filip Chindea
Privind \( \pmod{p} \), \( p | q + 7 \). In cazul \( q + 7 \ge 2p \), \( p \ge 7 \) obtinem \( q \ge 2p - 7 \ge p \), deci \( p^q \ge q^p \) si astfel \( 2p^q - q^p \ge 9 \), cum \( p \ge 3 \) (clasic - analizam functia \( x^{1/x} \)). Cazurile \( p \in \{2, 3, 5\} \) se arata, prin inductie simpla, ca nu furnizeaza solutii cu exceptia \( (p, q) \in \{(2, 3); (5, 3)\} \).
Pentru \( p = q + 7 \), setam \( s := q/7 \) si deci \( 2 \cdot 7^{7s} \cdot (s + 1)^{7s} - 7^{7s + 7} \cdot s^{7s + 7} = 7 > 0 \), si astfel \( s^7 < 2/7^7 \cdot \left((1 + 1/s)^s\right)^7 < 2 \cdot (e/7)^7 \), \( s \le \sqrt[7]{2} \cdot e / 7 \le ( 1,2 \cdot 2,8 )/7 < 4/7 \), \( q \le 3 \) care nu furnizeaza nicio solutie.

PS. Original: vad ca daca scriu in TEX "8" urmat de ")" apare \( 8) \)