polynomials

[mohlam12]

Hi everyone,

I have to demonsrate that for every real polynomial, P Q and R, I have :

P²=X(Q²+R²) ==imply==> P=Q=R=0

Using degrees, we can easily demonsrate the above. However, I'm looking for another way, without using that.

[Tom Mattson]

What is X? :confused:

[mohlam12]

X is the variable ! We can rewrite the problem this way :
if P(X), Q(X) and R(X) belongs to R[X], then writing P²(X) = X(Q²(X) + R²(X)) means P(X)=Q(X)=R(X)=0

Someone adviced me to start with demonstrating that constant coefficients are equal to 0 ? :s

[HallsofIvy]

The problem still makes no sense. Surely you don't mean "for every real polynomial, P Q and R" and then say "P= Q= R= 0".

Suppose Q= R= x and P= 2x3. Does that contradict what you are trying to prove? Are you assuming that P, Q, and R are have the same degree?

[mohlam12]

My bad ! Here's the exact question :
"Demonstrate that if P, Q, and R belong to R[X], therefore P² - XQ² = XR² imply that P=Q=R=0"

Big logic mistake in my first post -.- sorry

[HallsofIvy]

I think the "degree argument" you mentioned is still the simplest and best way to go. If n is the higher degree of Q and R, then the degree of xQ2+ R2 is 2n+ 1 so the degree of P would have to be (2n+1)/2, not an integer.

[Hurkyl]

At the risk of pointing out the obvious, f(x)^2 \geq 0 for every real x....