(k + 3 ) ∈ D5   k+3 ∈ {-5; -1; 1; 5}    k ∈ {-8; -4; -2; 2} (3k - 1) ∈ D5   3k-1 ∈ (-5; -1; 1; 5;}  3k ∈ {-4; 0; 2; 6}     k ∈ {0; 2} {-8; -4; -2; 2} ∩ {0; 2}  = 2   k = 2

x   si   y    trebuie  sa   fie   multiplii   de   5 Pt   k>0 U(x)={0,5} U(k+3)=0=>k1={7,17,27...} U(k+3)=5  =>k2={2,12,22,...} Pui   aceleasi   conditii   pt  U(y) U(3k-1)=0  K3={7,17,27...} U(3k-1)=5 =>  U(3k)=6  => k4={2,12,22...} Determini   k1Uk2   si   K3 UK4   Le   iyersectezi   si   obtii [K1Uk2]∩[K3UK4]={7,17,27...}U{2,12,22,...} (A revin  imediat Cazul   k<0 U(k+3)=5  => k5={-8,-18,...] U(k+3)=0 => k6={-3,-13,-23,...} U(3k-1)=0  k7={-3,-13,-23....} U(3k-1)=5  =<  k8=(-8,-18,-28...} [k5Uk6]∩{k7UK8[={-8,-18,-28,...}U{-3.-13,-23...} B{ In   final   faci  AUB