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Consider the recurrence $(a,b,c)\mapsto \left(\frac{3a+4b}{5}, \frac{4a-3b}{5}, c\right)$ and suppose that we start with $(2,3,2)$, and allow swapping positions before applying the recurrence again.

If we start with $(2,3,2)$, will we be able to obtain $(4,2,0)$?

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    @Arturo Magidin Thanks for the edit.2012-04-10

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Hint: Let $f(A,B,C)=A^2+B^2+C^2$. Note that the transformations leave $f$ invariant. You can interpret this geometrically (the distance to the origin does not change), or purely algebraically.

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    @MaoYiyi: The point is that each idea that you learn in solving problems adds to **your** knowledge.2012-04-10