Ecuatie URSS 1975
Posted: Sun Aug 09, 2009 9:05 pm
Sa se rezolve ecuatia :
\( 1!+2!+3!+...+(x+1)!=y^{z+1}\ , \ x,y,z\in \mathbb{N} \)
\( 1!+2!+3!+...+(x+1)!=y^{z+1}\ , \ x,y,z\in \mathbb{N} \)
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.