Ecuatie URSS 1975

alex2008 · Post by **alex2008** » Sun Aug 09, 2009 9:05 pm

Sa se rezolve ecuatia :

\( 1!+2!+3!+...+(x+1)!=y^{z+1}\ , \ x,y,z\in \mathbb{N} \)

Theodor Munteanu · Post by **Theodor Munteanu** » Mon Sep 21, 2009 4:34 pm

Observam ca avem o infinitate de triplete (1,1,n).
Daca \( x \geq 8 \) si \( z\geq 2 \) ecuatia nu are solutii deoarece 1!+2!+...+8!=46233 care se divide doar la 9, iar celalalt membru se divide cel putin la 27 in cazul in care presupunem ca y=3k (altfel evident nu gasim solutii).
Daca luam \( x \leq 7 \) prin incercari observam ca nu are solutii.
Daca \( x \geq 8 \) si z=0 ec are o infinitate de solutii.
Daca \( x \geq 8 \) si z=1 1!+2!+3!+4!=33 deci \( y^2=M5+3 \) imposibil iar 1+2!+3!=9=\( 3^2 \) deci (2,3,1) e o alta solutie a ecuatiei.

alex2008 · Post by **alex2008** » Mon Sep 21, 2009 7:57 pm

\( (3,3,1) \) nu e solutie.

Laurentiu Tucaa · Post by **Laurentiu Tucaa** » Mon Sep 21, 2009 8:42 pm

Cred ca a vrut sa spuna (2,3,1), care este solutie.