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{...--prophetes.ai

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

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
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.