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Question Number 197034 by Mingma last updated on 06/Sep/23

	Answered by TheHoneyCat last updated on 08/Sep/23
	$let f: { (R,→,R_+ ),(α, ,((4α^2 )/(1+4α^2 ))) :} if (x,y,z) is a “good” triplet then f(x)=y f(y)=z f(z)=x we notice that therefore, using f each triplet can be unicly defined by it′s first elelment x, the others beeing f(x) and f^2 (x):=f(f(x)) Thus there are as many triples as there are values x such that f^3 (x)=x f(x)=((4x^2 )/(1+4x^2 )) f^2 (x)=((4(((4x^2 )/(1+4x^2 )))^2 )/(1+4(((4x^2 )/(1+4x^2 )))^2 )) =((4(4x^2 )^2 )/((1+4x^2 )^2 +4(4x^2 )^2 )) =((2^6 x^4 )/(1+2^3 x^2 +2^6 x^4 )) f^3 (x)=((2^6 (((4x^2 )/(1+4x^2 )))^4 )/(1+2^3 (((4x^2 )/(1+4x^2 )))^2 +2^6 (((4x^2 )/(1+4x^2 )))^4 )) =((2^(14) x^8 )/((1+4x^2 )^4 +2^7 (1+4x^2 )^2 x^4 +2^(14) x^8 )) =((2^(14) x^8 )/(1+2^4 x^2 +224x^4 +1280x^6 +18688x^8 )) f^3 (x)=x has 3 solutions... x=(1/2) (I let you check it) x=0 (I let you check it) and one last point that I was not able to compute algebraicaly. (actualy I was, but we shall see that later) you can verify that there are only 3 because: f^3 (x)≥0 ⇒x≥0 f^3 (x) is strictly monotonous over R_+ between 0 and 1/2 ∃y∣f^3 (y)<y between 1/2 and 0.56 ∃y∣f^3 (y)>y beetween 0.57 and ∞ ∃y∣f^3 (y)<y this gives us exactly 3 values for x and therefore exactly 3 triples.$
	$let$ $f : {\begin{cases} R & \to & R_{+} \\ α & \frac{4 α^{2}}{1 + 4 α^{2}} \end{cases}$ $if (x, y, z) is a ‘ ‘ {good}^{″} triplet then$ $f (x) = y$ $f (y) = z$ $f (z) = x$ $we notice that therefore, using f$ $each triplet can be unicly defined by {it}^{'} s$ $first elelment x, the others beeing f (x) and$ $f^{2} (x) := f (f (x))$ $Thus there are as many triples as there are$ $values x such that f^{3} (x) = x$ $f (x) = \frac{4 x^{2}}{1 + 4 x^{2}}$ $f^{2} (x) = \frac{4 {(\frac{4 x^{2}}{1 + 4 x^{2}})}^{2}}{1 + 4 {(\frac{4 x^{2}}{1 + 4 x^{2}})}^{2}}$ $= \frac{4 {(4 x^{2})}^{2}}{{(1 + 4 x^{2})}^{2} + 4 {(4 x^{2})}^{2}}$ $= \frac{2^{6} x^{4}}{1 + 2^{3} x^{2} + 2^{6} x^{4}}$ $f^{3} (x) = \frac{2^{6} {(\frac{4 x^{2}}{1 + 4 x^{2}})}^{4}}{1 + 2^{3} {(\frac{4 x^{2}}{1 + 4 x^{2}})}^{2} + 2^{6} {(\frac{4 x^{2}}{1 + 4 x^{2}})}^{4}}$ $= \frac{2^{14} x^{8}}{{(1 + 4 x^{2})}^{4} + 2^{7} {(1 + 4 x^{2})}^{2} x^{4} + 2^{14} x^{8}}$ $= \frac{2^{14} x^{8}}{1 + 2^{4} x^{2} + 224 x^{4} + 1280 x^{6} + 18688 x^{8}}$ $f^{3} (x) = x has 3 solutions . . .$ $x = \frac{1}{2} (I let you check it)$ $x = 0 (I let you check it)$ $and one last point that I was not able to$ $compute algebraicaly . (actualy$ $I was, but we shall see that later)$ $you can verify that there are only 3 because :$ $f^{3} (x) ⩾ 0 \Rightarrow x ⩾ 0$ $f^{3} (x) is strictly monotonous over R_{+}$ $between 0 and 1 / 2 \exists y ∣ f^{3} (y) < y$ $between 1 / 2 and 0.56 \exists y ∣ f^{3} (y) > y$ $beetween 0.57 and \infty \exists y ∣ f^{3} (y) < y$ $this gives us exactly 3 values for x$ $and therefore exactly 3 triples .$
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