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Let $x, y$ be real numbers such that the numbers $x+y,x^2+y^2$ and $x^3+y^3$ are integers. Prove that $x^n+y^n$ is an integer for all natural $n$.

My attempt: Let $x+y=l, x^{2}+y^{2}=m$ for integers $l,m$, then $x y=\frac{l^2-m}{2}$. Now, $\begin{aligned} x^3+y^3 &=(x+y)^3-3 x y(x+y) \\ &=l^3-3 \cdot\left(\frac{l^2-m}{2}\right) \cdot l \\ &=\frac{3 m l}{2}-\frac{l^3}{2} \\ &=\frac{1}{2}\left[3 m l-l^3\right] \end{aligned}$ which is an integer, but i don't know what to do next.

Bill Dubuque
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轻型八神
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    @KCd well because that's how it was given in the original question, and I didn't change anything of it while posting it. Also how's this relevant at all to the problem anyway? Having parentheses or not isn't going to change anything right? – 轻型八神 Dec 01 '22 at 09:32
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    Hint: for a counterexample let $y = -x$. It seems the problem was copied incorrectly - it should be $x^4+y^4,,$ not $x^3+ y^3,,$ e.g. see AoPS (link found by John Omielan by an Approach0 search in a deleted answer below). – Bill Dubuque Dec 01 '22 at 11:10
  • Iirc the prior linked 2008 AoPS proof (post #4) also appears here in a number of places, but I could only find one such post in 2017 via a quick search. But possibly I am recalling older posts on sci.math or Ask an Algebraist. – Bill Dubuque Dec 01 '22 at 12:35

1 Answers1

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This is false. Take $x=\dfrac{1}{\sqrt{2}}$ and $y=-\dfrac{1}{\sqrt{2}}$.

Then $x+y=0,x^2+y^2=1,x^3+y^3=0$ but $x^4+y^4=\dfrac{1}{2}$.

GreginGre
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