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Exercise. Let $a,b$ be real numbers, such that $a+b$ and $ab$ are integers. Show that $\forall n\in \Bbb N$: $(a^n+b^n)\in \Bbb Z$.

I tried to use the binomial formula: $(a+b)^n=\sum_{k=0}^n{{n}\choose{k}}a^{n-k}b^k$, saying that since $a+b$ is an integer, then $(a+b)^n$ is also an integer. But I could not say anything about members of the sum because of the different powers of $a$ and $b$. After that, I tried to determine the values of $a$, and $b$, failing. Finally, I found a way to show that $a^{2^n}+b^{2^n}$ is an integer, which is better than nothing but is not a solution to the exercise.

I would appreciate any help you can provide.

Bill Dubuque
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Lord Mansfield
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2 Answers2

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Proceed by strong induction by noting that $a^{n+1}+b^{n+1}=(a^n+b^n)(a+b)-ab(a^{n-1}+b^{n-1})$ and that $a^2+b^2=(a+b)^2-2ab$.

Vishnu Kadiri
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Here's a hint:

$(a^n + b^n)(a+b) = a^{n+1} + a^{n}b + b^{n}a + b^{n+1} = a^{n+1} + b^{n+1} + ab(a^{n-1}) + ab(b^{n-1})$

So:
$a^{n+1} + b^{n+1} = (a^n + b^n)(a+b) - ab(a^{n-1} + b^{n-1})$

Domenic Petzinna
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