find all polynomial $p(x)$ that satisfy the equation $ p^2(x)+p^2(x-1)+1 = \left(p(x)-x\right)^2$
Tìm tất cả các đa thức $P(x)$ thoả mãn $$ P^2(x)+P^2(x-1)+1 = \left(P(x)-x\right)^2$$
Bắt đầu bởi stuart clark, 13-03-2011 - 13:56
#1
Đã gửi 13-03-2011 - 13:56
#2
Đã gửi 27-03-2011 - 10:26
I'm not sure
I I think so:
This equation is equivalent to $P^2(x-1)+2P(x)+1=x^2$
Let n=degree of $P(x)$
If n=0, then $P(x)=a$. Then LHS has degree $\leq 1$ while RHS has degree 2, hence no solution
If n=1, then $P(x)=ax+b$
Identification Then find a, b
If $n> 1$. Then LHS has degree $>2$ while RHS has degree 2, hence no solution
@stuart clark: Welcome to VMF
I I think so:
This equation is equivalent to $P^2(x-1)+2P(x)+1=x^2$
Let n=degree of $P(x)$
If n=0, then $P(x)=a$. Then LHS has degree $\leq 1$ while RHS has degree 2, hence no solution
If n=1, then $P(x)=ax+b$
Identification Then find a, b
If $n> 1$. Then LHS has degree $>2$ while RHS has degree 2, hence no solution
@stuart clark: Welcome to VMF
Bài viết đã được chỉnh sửa nội dung bởi SLNA: 27-03-2011 - 10:45
#3
Đã gửi 27-03-2011 - 10:51
I agree with you about degree of $P(x)$ equal to 1,but I think with the theory,we must find out value of $a,b$I'm not sure
I I think so:
This equation is equivalent to $P^2(x-1)+2P(x)+1=x^2$
Let n=degree of $P(x)$
If n=0, then $P(x)=a$. Then LHS has degree $\leq 1$ while RHS has degree 2, hence no solution
If n=1, then $P(x)=ax+b$
Identification Then find a, b
If $n> 1$. Then LHS has degree $>2$ while RHS has degree 2, hence no solution
@stuart clark: Welcome to VMF
"Do you still... believe in me ?" Sarah Kerrigan asked Jim Raynor - Starcraft II:Heart Of The Swarm.
#4
Đã gửi 27-03-2011 - 11:02
Yes
I think it has quite a lot of value
I think it has quite a lot of value
Bài viết đã được chỉnh sửa nội dung bởi SLNA: 27-03-2011 - 11:04
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