2017/007) if $\alpha,\beta$ are the roots of $x^2-5x+1$ show that $\alpha^n+\beta^n$ is an integer and not divisible by 4 for any integer n

we have for n = 1 $\alpha+\beta = 5 $ which is integer not divsible by 4
and $\alpha\beta=1$
hence $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 5^2 - 2 = 23$ integer and not divisible by 4
hence it is true for n =1 and 2
let it be true for n = 1 to k for k >=2
now
$( \alpha^n+\beta^n)(\alpha+\beta) = \alpha^{n+1}+\beta^{n+1} + (\alpha^{n-1} + \beta^{n-1} \alpha\beta$
or $\alpha^{n+1}+\beta^{n+1} = 5 (\alpha^n+\beta^n) - (\alpha^{n-1} + \beta^{n-1}$
or $\alpha^{n+1}+\beta^{n+1} =  4(\alpha^n+\beta^n) +  (\alpha^n+\beta^n)- (\alpha^{n-1} + \beta^{n-1} \alpha\beta$
Now let n be smallest integer so that $\alpha^{n}+\beta^{n}$ is divisible by 4
so $\alpha^{n-1}+\beta^{n-1} - \alpha^{n-2}+\beta^{n-2}$ is divsible by 4
so $\alpha^{n-3}+\beta^{n-3} = alpha^{n-1}+\beta^{n-1} - \alpha^{n-2}+\beta^{n-2} - 4alpha^{n-2}+\beta^{n-2}$
is divisible by 4 which is contadiction.
hence proved