Am cautat sa rescriu inegalitatile in n variante.Tot n-am reusit. As putea presupune ca m<n<n, pt m,n,p numere prime distincte! Dar…??? Oare-i ok? Poate ma ajuta cineva!
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Solutia -unica – este 2,3,5. Ramine de demonstrat …
O idee: Daca a = 2 si n = 3, inlocuind se obtine 6p-11 care pentru p = 5 da valoarea 19. Pentru orice p mai mare decit 5, rezultatul expresiei este mai mare decit 19.
Relatia fiind simetrica in m, n, p s-ar putea – eventual – arata ca pentru orice m si n mai mari decit 3 si 5 si p mai mare decit 7 nu obtinem solutii, deci (2,3,5) este singura solutie.
gm
Ai rezolvare in gm11
Tie-ti place rezolvarea dea acolo?! E prea aiurea!!! De ce sa mai imparti la mnp daca poti presupune de la inceput ca numerele sunt 2,3,5?
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