totul se rezolva pornind de la relatia: a^2 - b^2=(a-b)(a+b) sau pe caz particular cu b=1, a^2-1=(a-1)(a+1), sau a=1, 1-b^2=(1-b)(1+b) a^2 inseamna a la puterea 2
a)
1-16^4=1-2^16=(1-2^8)(1+2^8)=(1-2^4)(1+2^4)(1+2^8)=
=(1-2^2)(1+2^2)(1+2^4)(1+2^8)=(1-2)(1+2)(1+2^2)(1+2^4)(1+2^8)=
= - (1+2)(1+2^2)(1+2^4)(1+2^8)
si vezi ca se reduce cu termenul din fata si in final e zero
b)
4^10 +1=2^20 + 1
16^10 +1=2^40 + 1
(2^10 + 1)(2^10 - 1)=2^20 - 1
(2^20 - 1)(2^20 + 1)=2^40 - 1
(2^40 - 1)(2^40 +1)=2^80 - 1
in final avem:
256^10 - (2^80 - 1) + 1=2^80 - 2^80 +1 +1=2
c)
facem mai intai pe -2005^8 + 1= -(2005^8 - 1)= -(2005^4 - 1)(2005^4+1)=
=-(2005^2 - 1)(2005^2 +1)(2005^4 +1)=
= -(2005 -1)(2005+1)(2005^2+1)(2005^4+1)=
= -2004*2006(2005^2+1)(2005^4+1)
si acum se observa ca se reduce cu primul termen si in final e zero
