USA TSTST 2014 #3
by Wolstenholme, Aug 1, 2014, 10:20 PM
Find all polynomials 
with real coefficients that satisfy ![\[P(x\sqrt{2})=P(x+\sqrt{1-x^2})\]](//latex.artofproblemsolving.com/4/2/b/42b73cb625e5bf5ff0c8ee898ebb0d7caecf0083.png)
for all real 
with 
.
Solution:
Consider the monic polynomial
of minimal degree that satisfies the equation. Now, assume WLOG that 
. Letting 
we have that 
. Letting 
we have that 
. Letting 
we have that 
. Letting 
we have that 
.
Now, differentiating both sides we have that
. Letting 
we have that 
. Letting we have that 
. Letting we have that 
.
This implies that
is a factor of 
and upon checking we find that this eight degree polynomial (with any real constant added to it) is the minimal solution to the equation.
It is also clear that any polynomial in is a solution. I claim that these are the only solutions. Consider an arbitrary polynomial in ![$ \mathbb{R}[x] $](//latex.artofproblemsolving.com/6/e/0/6e04cf86b84b0188e2268bbdc853f537c7b684b4.png)
that is a solution to this equation. After subtracting off its constant term, it is clear that this polynomial must be a multiple of or else we could reduce it modulo and contradict the minimality of . After subtracting the constant term and dividing by , we get a new polynomial that also is a solution to the equation. After repeating this process, we inductively obtain the desired result.
Interestingly, this was the problem on the entire TSTST that took me the least time. This year's TSTST and USAMO were weird for me... I consider Geo and NT to be my best subjects but I did not get a single Geo or NT question right on either of the two tests, and got every single Combo and Alg right on both tests
with real coefficients that satisfy ![\[P(x\sqrt{2})=P(x+\sqrt{1-x^2})\]](http://latex.artofproblemsolving.com/4/2/b/42b73cb625e5bf5ff0c8ee898ebb0d7caecf0083.png)
for all real 
with 
.Solution:
Consider the monic polynomial
of minimal degree that satisfies the equation. Now, assume WLOG that 
. Letting 
we have that 
. Letting 
we have that 
. Letting 
we have that 
. Letting 
we have that 
.Now, differentiating both sides we have that
. Letting 
we have that 
. Letting we have that 
. Letting we have that 
.This implies that
is a factor of 
and upon checking we find that this eight degree polynomial (with any real constant added to it) is the minimal solution to the equation.It is also clear that any polynomial in
is a solution. I claim that these are the only solutions. Consider an arbitrary polynomial in ![$ \mathbb{R}[x] $](http://latex.artofproblemsolving.com/6/e/0/6e04cf86b84b0188e2268bbdc853f537c7b684b4.png)
that is a solution to this equation. After subtracting off its constant term, it is clear that this polynomial must be a multiple of or else we could reduce it modulo and contradict the minimality of . After subtracting the constant term and dividing by , we get a new polynomial that also is a solution to the equation. After repeating this process, we inductively obtain the desired result.Interestingly, this was the problem on the entire TSTST that took me the least time. This year's TSTST and USAMO were weird for me... I consider Geo and NT to be my best subjects but I did not get a single Geo or NT question right on either of the two tests, and got every single Combo and Alg right on both tests
so I couldn't get the 28 I ""deserved." So yes I count it as I got this problem.
August 2016
June 2016