Problem 2

Let

P_1(x) = x^2 - 2

and

P_j(x) = P_1(P_{j-1}(x))

for

j = 2, 3, \ldots

. Prove that, for any positive integer

n

, the roots of the equation

P_n(x) = x

are all real and distinct.

Step 4 of 6: Solve the trigonometric equation

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In plain words

The two sign choices $\pm$ inside the general cosine-equality identity are exactly what produce two separate arithmetic-progression families of angles, which will be counted separately.

\theta = \frac{2k\pi}{2^n-1} \ \text{ or } \ \theta = \frac{2l\pi}{2^n+1}

Detailed analysis

$\cos A=\cos B$ holds exactly when $A=\pm B+2k\pi$ for some integer $k$ . Applying this with $A=2^n\theta$ , $B=\theta$ gives two families: $(2^n-1)\theta=2k\pi$ , i.e. $\theta=\dfrac{2k\pi}{2^n-1}$ ; or $(2^n+1)\theta=2l\pi$ , i.e. $\theta=\dfrac{2l\pi}{2^n+1}$ , for integers $k,l$ .