Răspuns:
36. (2n-1) | (4n+5)
dar (2n-1) | (2n-1)
⇔ (2n-1) | (4n+5)
(2n-1) | 2x(2n-1)
Atunci 2n-1 divide si diferenta
⇔ (2n-1) | (4n+5-4n+2)
⇔ (2n-1) | 7⇒ 2n-1∈D7
2n-1∈{-7,-1,1,7}
2n∈{-6,0,2,8}
⇒n∈{-3,0,1,4}
dar n∈Ν
Din cele 2 ecuatii⇒ n∈{0,1,4}
Analog la celelalte
