Integers whose sum and product are integers Let $a$, $b$ be real numbers such that $a + b$ and $ab$ are integers.
a. Prove that $a^n + b^n$ is an integer for every natural number $n$.
b. Suppose that $a$ does not equal $b$. Prove that $\dfrac{a^n - b^n}{a - b}$ is an integer for every positive integer $n$.
Attempts
I am thinking mathematical induction can be applied to part a.
1) Base Case: n = 1
$a^1$ + $b^1$ = $a + b$ which is an integer.
2) Assume $a^{n-1}$ + $b^{n-1}$ is an integer. We want to show that $a^n$ + $b^n$ is also an integer.
3) $a^n$ + $b^n$ = ... Not sure what to do next

Mathematical induction seems to also apply to part b.
1) Base Case: n = 1
$\dfrac{a^1 - b^1}{a - b}$ = $\dfrac{a-b}{a - b}$ = 1 which is an integer.
2) Assume $\dfrac{a^{n-1} - b^{n-1}}{a - b}$  is an integer. We want to show that $\dfrac{a^{n} - b^{n}}{a - b}$ is also an integer.
3) $\dfrac{a^{n} - b^{n}}{a - b}$ = ... Not sure what to do next
 A: a. As you started, we use induction on $n$. The case where $n=1$, holds. Assume that it also holds for all $n\leq k$ (strong induction). We have
$$a^{k+1}+b^{k+1}=(a+b)(a^k+b^k)-ab(a^{k-1}+b^{k-1})$$
which is integer, as it is products and sums of integers. So the proof is complete.
b. Notice that
$$\frac{a^n-b^n}{a-b}=a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}$$
For $n$ even you can rearrange the sum above, in factors of the tipe $(ab)^j(a^k+b^k)$, which is integer from a. For $n$ odd, you can still rearrenge the sum in a similar way, with the difference that the middle term is $(ab)^{\frac{n-1}{2}}$, which is also integer.
A: Induction is a good idea; I would look at $(a+b)^n$ using the Binomial Theorem, do some rearrangement, and apply the inductive hypothesis.
For part b, notice that
$$a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b + a^{n-3}b^2 + \ldots + b^{n-1}).$$
The result from part a might be helpful.
A: For a) 
$\\ { a }^{ 2 }+{ b }^{ 2 }={ \left( a+b \right) \left( a+b \right)  }-2ab,\quad integer\\ { a }^{ 3 }+{ b }^{ 3 }=\left( a+b \right) { \left( { a }^{ 2 }+{ b }^{ 2 } \right)  }-ab\left( a+b \right) ,\quad integer\\ { a }^{ 4 }+{ b }^{ 4 }=\left( a+b \right) { \left( { a }^{ 3 }+{ b }^{ 3 } \right)  }-ab\left( { a }^{ 2 }+{ b }^{ 2 } \right) ,\quad integer\\ \vdots \\ { a }^{ n }+{ b }^{ n }=\left( a+b \right) { \left( { a }^{ n-1 }+{ b }^{ n-1 } \right)  }-ab\left( { a }^{ n-2 }+{ b }^{ n-2 } \right) ,\quad integer$ 
