Pell equations and Chebyshev polynomials
by math154, Aug 5, 2011, 4:15 PM
Hmm...
(MOP 2011 Handout.) Prove that for any given positive integer
, there exists a unique polynomial 
of degree with integer coefficients such that 
and 
is odd.
Edit: This is from China TST 2007 apparently.
As for actually using Chebyshev polynomials in number theory... well it turns out that
is almost never a perfect square for integers 
.
Edit: Alternatively, for the Pell equations, here's something David showed me: induct on
. Clearly 
is the only working pair in ![$\mathbb{R}[x]$](//latex.artofproblemsolving.com/a/8/8/a88b9f4858016b1771635c611d44b56161a08009.png)
for 
. Now suppose 
and 
for some 
, 
. Comparing 
coefficients yields 
, so 
. But 
(get this by considering 
) then reduces degree so 
generates all solutions up to sign.
(MOP 2011 Handout.) Prove that for any given positive integer
, there exists a unique polynomial 
of degree with integer coefficients such that 
and 
is odd.
,![$(x+1)f(x)^2-1=-[(-x+1)f(-x)^2-1]$](http://latex.artofproblemsolving.com/5/6/f/56f892adae2eb19f3babc43d9ae4b39527b2e68e.png)
rewrites as ![$x[p(x)+q(x)]^2+(1-x)p(x)^2=1$](http://latex.artofproblemsolving.com/0/f/1/0f17f3b994cd6058e0d74a36f0e02117c874e557.png)
.
.![\[r(x)[r(x)+2xr'(x)]=p(x)[p(x)+2(x-1)p'(x)].\]](http://latex.artofproblemsolving.com/a/4/4/a44b218932f91be772a467a188e35e4c13a2f3e6.png)
Clearly 
or else (*) can't hold (the leading coefficients must be the same up to sign), so because 
,
must divide 
and so there exists a constant 
such that ![$p(x)=c[r(x)+2xr'(x)]$](http://latex.artofproblemsolving.com/a/0/e/a0ea1326f304951b6b5137bd4fe9e99eb1a84400.png)
.![\[1=[x+c^2(1-x)]r(x)^2+4c^2x(1-x)r'(x)r(x)+4c^2x^2(1-x)r'(x)^2.\]](http://latex.artofproblemsolving.com/8/e/7/8e77e4dd72d7062cda2c7dde7d2134c9723950e5.png)
From the quadratic formula, we find![\[r'(x)=-\frac{c(x-1)r(x)\pm\sqrt{(x-1)[xr(x)^2-1]}}{2cx(x-1)},\]](http://latex.artofproblemsolving.com/e/1/9/e19113c5e74680d44ca7670d4e065b1de4e3f7b5.png)
or letting 
so that 
and simplifying,![\[\frac{2cs'(x)}{\sqrt{s(x)^2-1}}=\sqrt{\frac{1}{(2x-1)^2-1}},\]](http://latex.artofproblemsolving.com/8/8/3/883eac0fd448800bd487952a800e2781e95bbf71.png)
where WLOG 
.![\[\int\frac{1}{\sqrt{x^2-1}}\;dx=\ln(x+\sqrt{x^2-1})+C,\]](http://latex.artofproblemsolving.com/d/9/5/d95f7b8b07dd7fc46f4ea4374f42eab0eadbd1db.png)
so for some constant 
we have![\[c\ln(s(x)+\sqrt{s(x)^2-1})=\ln(\sqrt{x}+\sqrt{x-1})+C_0,\]](http://latex.artofproblemsolving.com/f/b/d/fbde214e732801b65841fa464eb52c90dd51cd20.png)
i.e.![\[\frac{1}{s(x)-\sqrt{s(x)^2-1}}=s(x)+\sqrt{s(x)^2-1}=e^{C_0/c}(\sqrt{x}+\sqrt{x-1})^{1/c},\]](http://latex.artofproblemsolving.com/b/3/b/b3babadcf9a114b76c5885670cda724efc7bc942.png)
so there exist constants 
such that![\[r(x)=\frac{e^{C_1}(\sqrt{x}+\sqrt{x-1})^{C_2}+e^{-C_1}(\sqrt{x}-\sqrt{x-1})^{C_2}}{2\sqrt{x}}.\]](http://latex.artofproblemsolving.com/4/d/a/4daf28c82169e1af2e21b94b9af2c7a83fade58f.png)
Since 
is a polynomial, we clearly have 
and thus 
.
(up to sign of course).
upon the substitutions 
,
.Edit: This is from China TST 2007 apparently.
As for actually using Chebyshev polynomials in number theory... well it turns out that
is almost never a perfect square for integers 
.Edit: Alternatively, for the Pell equations, here's something David showed me: induct on
. Clearly 
is the only working pair in ![$\mathbb{R}[x]$](http://latex.artofproblemsolving.com/a/8/8/a88b9f4858016b1771635c611d44b56161a08009.png)
for 
. Now suppose 
and 
for some 
, 
. Comparing 
coefficients yields 
, so 
. But 
(get this by considering 
) then reduces degree so 
generates all solutions up to sign.This post has been edited 6 times. Last edited by math154, Feb 28, 2013, 5:35 PM
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