x+1/x=a∈Z
(x+1/x)²=a²∈Z
x²+1/x²+2=a²∈Z
x²+1/x²=a²-2∈Z
(x+1/x)³=(x+1/x)²(x+1/x)=(a²-2)*a∈Z ca produsde 2 numere intregi
(x+1/x)^4=a^4∈Z
x^4+4x³*1/x+6 x²*1/x²+4x*1/x³+1/x^4=a^4
x^4+4(x²+1/x²)+6+1/x^4=a^4
x^4 +4(a²-2)+6+1/x^4=a^4
x^4+1/x^4=a^4-6-4(a²-2)∈Z
presupunem
x^n+1/x^n ∈Z
(x+1/x)^(n+1)==a^(n+1)∈Z=
x^(n+1)+Comb de n+1 luate cate 1*x^n * 1/x+Combde n+1 luate cate 2 x^(n-1)*1/x²+.....................Comb de n+1 luate cate n+1* 1/x^(n+1)
in rest vezi atas
am demonstrat prin inductie ca x^n+1/x^n ∈Z
punctul b) nu am reusit, v,foaia a doaua
mai trebuie sa arat ca (x+1/x)^2016 este par
adica x+1/x par
presyupunem x+1/x impar
(x²+1)/x
