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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 $(4,2,0)$?

Arturo Magidin
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yiyi
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1 Answers1

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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.

André Nicolas
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  • I sort of see how it would be invariant, but I am not getting the idea of $f(A,B,C) = A^2+B^2+C^2$. – yiyi Apr 10 '12 at 03:49
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    You might recognize that your transformation is a rotation by $\arccos(\frac{4}{5})$ followed by a reflection in $y=x$. Thus, the distance to $O$ never change.... – N. S. Apr 10 '12 at 03:56
  • Let $T$ be anyone of your allowed transformations (the main one, or any swap). Let $(A_1,B_1,C_1)=T(A,B,C)$. You can check easily that $A_1^2+B_1^2+C_1^2=A^2+B^2+C^2$. This is trivial if $T$ is a swap. For the main transformation, it is a calculation, since $\left(\frac{3A+4B}{5}\right)^2+\left(\frac{4A-3B}{5}\right)^2+C^2=A^2+B^2+C^2$. Originally, your point is on a sphere centre the origin, radius $\sqrt{17}$, and under the transformations it stays on that sphere. But $(4,2,0)$ is not on that sphere! – André Nicolas Apr 10 '12 at 03:57
  • @AndréNicolas I bow to your knowledge. – yiyi Apr 10 '12 at 03:59
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    @MaoYiyi: The point is that each idea that you learn in solving problems adds to your knowledge. – André Nicolas Apr 10 '12 at 04:02