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Found this interesting problem online:

Let $T_n(x)$ be the $n$-th Chebyshev polynomial. Show that for $n$ odd the polynomial $$T_{n}(x)^2 + T_n(y)^2-1$$ is divisible by $x^2 + y^2-1$.

Notes:

  1. It checks for all values I've tried.

  2. There is a recurrence for $a_n\colon = T_{2n+1}(x)^2$ as follows

$$a_{n+3} - (16 x^4 - 16 x^2 + 3) a_{n+2} + (16 x^4 - 16 x^2 + 3) a_{n+1} - a_n = 0$$

obtained from the recurrence $T_{n+2}(x) - 2 x T_{n+1}(x) + T_n(x) = 0$. This may allow us to use induction but it's not so clear.

Any feedback would be appreciated!

orangeskid
  • 53,909

3 Answers3

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Let $\,x = \cos\alpha, y=\cos\beta, p = \cos (\alpha-\beta), q = \cos(\alpha+\beta)\,$. It should be noted that the elementary symmetric polynomials in $\,p,q\,$ can be written as polynomials in $\,x,y\,$:

$$ \begin{align} p + q &= \cos(\alpha-\beta)+\cos(\alpha+\beta) = 2 \cos\alpha \cos\beta = 2 xy \\ pq &= \cos(\alpha-\beta)\cos(\alpha+\beta) = \cos^2\alpha \cos^2\beta - \sin^2\alpha \sin^2\beta = x^2y^2 - (1-x^2)(1-y^2) \end{align} $$

Then:

$$ \begin{align} x^2 + y^2 - 1 &= \cos^2(\alpha)+\cos^2(\beta)-1 \\ &= \frac{1}{2}\left(\cos(2\alpha) + \cos(2\beta)\right) \\ &= \cos(\alpha-\beta)\cos(\alpha+\beta) \\ &= pq \\ T_{n}(x)^2 + T_n(y)^2-1 &= \cos^2(n\alpha) + \cos^2(n\beta) - 1 \\ &= \cos\left(n(\alpha-\beta)\right)\cos\left(n(\alpha+\beta)\right) \\ &= T_n(p) T_n(q) \end{align} $$

Chebyshev polynomials of odd order $\,n\,$ are known to be odd, so $\,T_n(p) = p \cdot R_n(p^2)\,$ for some polynomial $\,R_n\,$, so it follows that:

$$ T_n(p)T_n(q) = pq \cdot R_n(p^2)R_n(q^2) $$

The second factor $\,R_n(p^2)R_n(q^2)\,$ on the RHS is a symmetric polynomial in $\,p,q\,$, which can be written as a polynomial in $\,p+q,pq\,$, which are in turn polynomials in $\,x,y\,$ per the earlier observation, so $R_n(p^2)R_n(q^2) = S_n(x,y)$ for some polynomial $S_n$. Putting it all together:

$$ T_{n}(x)^2 + T_n(y)^2 - 1 = T_n(p)T_n(q) = pq \cdot R_n(p^2)R_n(q^2) = (x^2 + y^2 - 1) \cdot S_n(x,y) $$

[ EDIT ] $\;$ The above proves the equality for $\,|x|, |y| \le 1\,$ because of the trigonometric substitution that was used. However, that is enough to prove the unrestricted polynomial identity, given that two bivariate polynomials that are equal on an open ball in $\,\mathbb R^2\,$ must be identical, see for example A polynomial that is zero on an open set.

dxiv
  • 76,497
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Unless I am grossly mistaken, you have done the hardest part ? Let $g(x,y) = x^2+y^2-1$ and $$x^2\equiv1-y^2\pmod g\implies x^4\equiv 1-2y^2+y^4\pmod g\implies $$ $$\equiv x^4-x^2\equiv y^4-y^2\pmod g.$$

In other words, let the coefficients $16x^4-16x^2+3\equiv 16y^4-16y^2+3 := X\pmod g$ and then strong induction works: $$a_{n+3}(x)+a_{n+3}(y) \equiv X\left(\left(a_{n+2}(x)+a_{n+2}(y)\right)-\left(a_{n+1}(x)+a_{n+1}(y)\right)\right) + \left(a_{n}(x) + a_{n}(y)\right)\equiv $$

$$\equiv X(1-1)+1=1\pmod g$$

dezdichado
  • 13,888
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We have $T_n(\cos \theta) = \cos n \theta$ for all $n$ natural, and also $T_n(\sin \theta) = (-1)^{\frac{n-1}{2}} \sin n \theta$ for all $n$ odd (use $\sin \theta = \cos ( \frac{\pi}{2}- \theta)$ ). Therefore for $n$ odd

$$T_n(\cos \theta)^2 + T_n(\sin \theta)^2 -1 = \cos^2 n \theta + \sin^2 n \theta -1 = 0$$

Now we only need to check that every polynomial $P(x,y)$ such that $P(\cos \theta , \sin \theta) \equiv 0$ is divisible by $x^2 + y^2-1$. Indeed, divide by $y^2 + x^2 -1$ with a remainder $A(x) y + B(x)$. We have $A(\cos \theta) \sin \theta + B(\cos \theta) \equiv 0$, so if $A$, $B$ were not both $0$ then $\sin \theta\equiv -\frac{B(\cos \theta)}{A(\cos\theta)}$, an even function of $\theta$, not possible.

orangeskid
  • 53,909