impartitorul g=x² -x +1 are Δ= -3 si radacinile x₁ = ( 1+ √3i) /2 si x₂=(1 -√3i)/2
care verifica ecuatia , g(x₁) =0 si g(x₂)=0 radacini care x₁³ = x₂³ = - 1
atunci gradul restului este 2 . r= ax+b
teorema impartirii cu rest
f= g·c +r
( x³₁ +x₁ +1) la 19 = ( x²₁ - x₁+1) · c₁ + ax₁+b
↓=-1 ↓=0
( x³₂+x₂+1 ) la 19=( x²₂ -x₂+1) · c₂+ax₂ +b
↓=-1
↓=0
(-1 +x₁+1)la 19= ax₁+b ax₁+b = x₁la puterea 19
(-1 +x₂+1) la 19=ax₂+b ax₂+b= x₂ la puterea 19
scadere a( x₁-x₂ ) = x₁laputrea 19 -x₂laputerea 19
√3i·a= (x₁ -x₂) ( x₁^18 + x₁^17x₂+x₁^16x₂² + ........+ x₁x₂^17+x₂^18)
↓ ↓ ↓
-1 x₁x₂·x₁^16 -1
↓ ↓
1 x₁ samd
√3i·a=√3i·(1-x₁+x₁²+1-x₁+x₁²+1-x₁+x₁²+x₁⁹x₂⁹+x₁⁸x₂¹⁰+x₁⁷x₂¹¹+x₁⁶x₂¹²+x₁⁵x₂¹³+x₁⁴x₂¹⁴+x₁³x₂¹⁵+x₁²x₂¹⁶+x₁x₂¹⁷+x₂¹⁸)=
=1-x₁+x₁²+1-x₁+x₁²+1-x₁+x₁²+1+x₂²-x₂+1+x₂²-x₂+1+x₂²-x₂+1=
=7 -3(x₁+x₂)+3( x₁²+x₂²) =7 -3·1+3(-1)=1 ⇒ a=1
b= 0
