5

Question:

prove or disprove :there exsit real numbers $a,b$ such follow two condition:

(1):$a+b$ is irrational

(2): for any postive integer $n\ge 2$, then $a^n+b^n$ is rational.

I have know if

$n=2k$ case is true,because I let $a=\sqrt{2}+1,b=\sqrt{2}-1$,so $$a^{2k}+b^{2k}=(\sqrt{2}+1)^{2k}+(\sqrt{2}-1)^{2k}\in Q$$

But for $n=2k+1$,I can't find a example.(if you can't find,can you prove when$n=2k+1$,there can't exsit?) Thank you for help

Bill Dubuque
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math110
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2 Answers2

8

Suppose that $a,b$ are real numbers such that $a^n+b^n$ is rational for all integers $n \ge 2$.

Since $a^2+b^2$ and $a^4+b^4$ are rational, $\dfrac{(a^2+b^2)^2-(a^4+b^4)}{2} = a^2b^2$ is also rational.

Then, since $(a^5+b^5)-(a^2+b^2)(a^3+b^3)+a^2b^2(a+b) = 0$, we have that

$\dfrac{(a^2+b^2)(a^3+b^3)-(a^5+b^5)}{a^2b^2} = a+b$ is rational, as desired.

EDIT: The above is only valid if $a^2b^2 \neq 0$, but the case where $a = 0$ or $b = 0$ is easy.

JimmyK4542
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5

Theorem. There are no real numbers $a,b$ such that $$\hbox{$a+b$ is irrational}$$ and $$\hbox{$a^2+b^2$, $a^3+b^3$, $a^4+b^4$ and $a^6+b^6$ are rational.}$$

Proof. Suppose that there are such $a,b$; clearly they are non-zero. Then $$2a^2b^2=(a^2+b^2)^2-(a^4+b^4)$$ and $$2a^3b^3=(a^3+b^3)^2-(a^6+b^6)$$ are rational, and so is $$ab=\frac{2a^3b^3}{2a^2b^2}\ .$$ Therefore $$(a+b)^2=(a^2+b^2)+2ab$$ is rational, and since $$(a+b)^3=(a^3+b^3)+3ab(a+b)$$ we have $$a+b=\frac{a^3+b^3}{(a+b)^2-3ab}$$ which is rational (note that with real $a,b$, the denominator of this last fraction cannot be zero).

David
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