I've solved the first part of the question in the following way(with the help of second hypothesis):

Given :x^2-(p+1)x+1=0 and alpha and beta are the roots .

so,alpha+beta = p+1 ..............(1)andalpha+beta is greater than or equal to 4.............(2){since it is given that p>=3}

alpha x beta =1 ................(3)

STEP1 -- To prove that P(1) is an integer.

alpha^1+beta^1 =alpha+beta -------> is an integer. [from (1)]

To prove that P(2) is an integer.

P(2)=alpha^2+beta^2=(alpha+beta)^2- 2alpha .beta

{(alpha+beta)^2 is >=16 [from 2] , 2alpha .beta =2 [from 3]

so, the value of alpha^2+beta^2=(alpha+beta)^2- 2alpha .beta becomes >=14 and hence it is an integer.}

STEP 2 INDUCTION ASSUMPTION

--------------------------------

Let P(k) be an integer.

so, alpha^k +beta^k be an integer.

Let P(k-1) be an integer.

so,alpha^(k-1) +beta^(k-1) be an integer.

Step3--To prove that P(k+1) is an integer.

P(k)=alpha^(k+1) +beta^(k+1)

=(alpha^k +beta^k )(alpha+beta) - alpha^k.beta-beta^k.alpha

=(alpha^k +beta^k )(alpha+beta) - alpha .beta{alpha^(k-1)+beta^(k-1)}

USING-

alpha+beta is greater than or equal to 4.............(2)

alpha x beta =1 ................(3)

(alpha^k +beta^k )(alpha+beta) >=4 [from above 2 relations], alpha^k +beta^k is an integer and alpha^(k-1) +beta^(k-1) is also an integer.

and alpha^(k-1)+beta^(k-1)

From induction assumption

Hence it has been proved that (alpha^k +beta^k )(alpha+beta) - alpha .beta{alpha^(k-1)+beta^(k-1)} is an integer.