Q: Using Binomial Theorem to solve a question The question is this:

"We assume that $a$,$b$ are positive numbers(they might not be integers)
  but  $a+b$ and $a*b$ are integers, prove that for every $n$ $\in$$\mathbb{N}$ the numbers $a^n + b^n$ are integers. 

Hint(Use Binominal Theorem).
I've tried to solve it but I couldn't. I could use some help

A: You don't need the Binomial Theorem. Use induction.
Note that $a^n+b^n$ holds for $n=0,1$ from the conditions provided. Now asssume it is true for $n=m$, $n=m+1$. Then using Newton's Identity $$a^{m+2}+b^{m+2}=(a+b)(a^{m+1}+b^{m+1})-ab(a^m+b^m)$$
We find it holds for $m+2$ as well.
Thus, our proof is done.
EDIT
Upon second consideration, one can prove this using the Binomial Theorem by using strong induction. Assume that the proposition holds true for all $n<m$.
If $m$ is even, $$a^m+b^m=(a+b)^m-\sum_{i=1}^{\frac{m-2}{2}} \binom{m}{i} a^ib^i(a^{m-i}+b^{m-i})-\binom{m}{\frac{m}{2}}a^{\frac{m}{2}}b^{\frac{m}{2}} \in \mathbb{Z}$$ By the inductive hypothesis.
If $m$ is odd, $$a^m+b^m=(a+b)^m-\sum_{i=1}^{\frac{m-1}{2}} \binom{m}{i} a^ib^i(a^{m-i}+b^{m-i})\in \mathbb{Z}$$
From the inductive hypothesis. We are done.

A: $$(a^n+b^n)(a+b)=a^{n+1}+b^{n+1}+ab(a^{n-1}+b^{n-1})$$
$$\iff a^{n+1}+b^{n+1}=(a^n+b^n)(a+b)-ab(a^{n-1}+b^{n-1})$$
Now use Strong Induction.
See : Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.
