A nice general way to reason about this problem for higher degree polynomials is to let Q(x) = x^2 -1 and rewrite the original equation: P(x^2 - 1) = P(x)^2 - 1 as P(Q(x)) = Q(P(x)) which can be written as P*Q = Q*P where * is function composition. Clearly P=Q is a solution. So is is P= Q*Q and P = Q*Q*Q and so on. So iterating Q gives all the solutions.
A mistake is made during evaluation of the quadratic polynomial. The constant equation should be a - b + c = c^2 -1. The c^2 term was inadvertently dropped. So the solution a = 1, b = 0, and c = -1 works and the polynomial solution is P(x) = x^2 - 1. This is clearly a solution of the original equation.
Why didn't we check any other form of function? (Logarithmic, trigonometric, exponential, mixed etc). We can state that P(x)=+-phi and P(x)=x are some of the solutions but not the only solution.
Because, as stated at the start of the video, this is a problem about polynomials. Once you introduce transcendental functions (or trig functions) you are no longer dealing with polynomials. The question in a comment further down about higher order polynomials still stands. I have no idea how to prove that.
In the left sides all powers are even, which means all powers on the right side should also be even which cant be true if P(x) has both even and odd degree terms. And then blablabla
Not really sure. The rank of p(x^2-1) is equal to the rank of (p(X))^2 as the leading coefficients are equal. We can show that the polynomials can only be even or odd but apart from that I would need a piece of paper X^2-1 even => P(x^2-1) => (P(X))^2-1 even => (P(X))^2-1=(P(-x))^2-1 => P(X)=P(-x) or P(X)=-P(-X) I see sybermath going the check route so we could check an even or odd polynomial too just to make sure no other solutions exist. Plug 1 in original expr P(0)=(P(1))^2-1 Plug √2 in : P(1)=(P(√2))^2-1 Plug in √(√2+1) P(√2)=(P(√(√2+1))^2-1 Does not seem conclusive but at least if P(X) is odd P(0)=0 which means P(1)=+1 or -1. Plugging 0 in P(-1)=-1 so P(1)=1 so the sum of coefficients for odd polynomial is 1. From Sybermath's lead coefficient always has to be 1 so the rest need to be 0or add to 0. Case rank=1 then P(X)=X verifies this condition checking original EQ x^2-1=x^2-1 true. Case rank=3 then P(X) =ax^3+bx and a+b=1 We need to derive another relation , but there are no nice integers to obtain integer values. Trying the original expression P(x^2-1)=(P(X))^2-1 a(x^2-1)^3+b(x2-1)=(ax^3+bx)^2-1 ax^6-3ax^4+3ax^2-a+bx^2-b=a^2x^6+2abx^4+b^2x^2-1 a^2=a and -3a=2ab and 3a+b=b^2 and -a-b=-1. Since a!=0 a=1 and b=0 but it does not verify relations in the middle In a general case P(X)=X*E where E is an even polynomial => (X^2-1)E(x^2-1)=x^2(E(X))^2-1 (X^2-1)E(x^2-1)=(x^2-1)(E(X))^2+(e(X))^2-1. So e(X) = 1*x^2m+sum(a2i * x^2i *(x^2-1)) a term that can factor out x^2-1 when dividing 1 and a part divisible by X^2-1 . This looks quite bad Going back to the odd polynomial if X=I then P(-2)=(P(I))^2-1 which has th sum of alternating sign coefficients inside the paranthesis P(2)=(P(√3))^2-1 which is the negated value of P(2) but each coefficient has different multiplier. Overall (P(√3))^2-1=-(P(I))^2+1 (P(√3)+P(I))(P(√3)-P(I))=2 I guess I am not clever enough for this Olympiad problem. This will get a relationship between coefficients but it won't be pretty Case p(X) even then P(0)=c X=1 P(0)=(P(1))^2-1=c X=0 P(-1)=P(1)=(P(0))^2-1 => P(1)=c^2-1 and (P(1))^2=c+1 (C^2-1)^2=c+1 C^4-2c^2+1=c+1 C(c^3-2c-1)=0 c(c+1)(c^2-c-1)=0 We have a few cases here: C=0 means p(1)=+1 or -1 the rank of polynomial is >0 so from Sybermath's observation the leading coefficient is +1 => P(X)=x^2k because the sum of coefficients is also 1
There are infinite solutions actually. You can iterate the polynomial x^2 +1 to get more solutions. See math stackexchange question 271337 for a discussion on this exact problem.
A nice general way to reason about this problem for higher degree polynomials is to let Q(x) = x^2 -1 and rewrite the original equation:
P(x^2 - 1) = P(x)^2 - 1
as
P(Q(x)) = Q(P(x))
which can be written as
P*Q = Q*P
where * is function composition.
Clearly P=Q is a solution.
So is is P= Q*Q and P = Q*Q*Q and so on. So iterating Q gives all the solutions.
A mistake is made during evaluation of the quadratic polynomial. The constant equation should be a - b + c = c^2 -1. The c^2 term was inadvertently dropped. So the solution a = 1, b = 0, and c = -1 works and the polynomial solution is P(x) = x^2 - 1. This is clearly a solution of the original equation.
oopsies!
2次方程式において、a=1 b=0 c=-1も成立するのでは？
Why didn't we check any other form of function? (Logarithmic, trigonometric, exponential, mixed etc). We can state that P(x)=+-phi and P(x)=x are some of the solutions but not the only solution.
Because, as stated at the start of the video, this is a problem about polynomials. Once you introduce transcendental functions (or trig functions) you are no longer dealing with polynomials. The question in a comment further down about higher order polynomials still stands. I have no idea how to prove that.
Try P(x) = x²-1
Nice - Happy Thanksgiving!
Same to you!
la même constatation pour c=-1pour la forme quadratique
In the left sides all powers are even, which means all powers on the right side should also be even which cant be true if P(x) has both even and odd degree terms. And then blablabla
Can you tell me in which order I should follow your channel playlists
Start with anything you like and then go from there
so how do we prove there are no other solutions for degrees > 2 ?
Not really sure. The rank of p(x^2-1) is equal to the rank of (p(X))^2 as the leading coefficients are equal. We can show that the polynomials can only be even or odd but apart from that I would need a piece of paper
X^2-1 even => P(x^2-1) => (P(X))^2-1 even => (P(X))^2-1=(P(-x))^2-1 => P(X)=P(-x) or P(X)=-P(-X)
I see sybermath going the check route so we could check an even or odd polynomial too just to make sure no other solutions exist.
Plug 1 in original expr P(0)=(P(1))^2-1
Plug √2 in : P(1)=(P(√2))^2-1
Plug in √(√2+1) P(√2)=(P(√(√2+1))^2-1
Does not seem conclusive but at least if P(X) is odd P(0)=0 which means P(1)=+1 or -1. Plugging 0 in P(-1)=-1 so P(1)=1 so the sum of coefficients for odd polynomial is 1. From Sybermath's lead coefficient always has to be 1 so the rest need to be 0or add to 0.
Case rank=1 then P(X)=X verifies this condition checking original EQ x^2-1=x^2-1 true.
Case rank=3 then P(X) =ax^3+bx and a+b=1
We need to derive another relation , but there are no nice integers to obtain integer values. Trying the original expression
P(x^2-1)=(P(X))^2-1
a(x^2-1)^3+b(x2-1)=(ax^3+bx)^2-1
ax^6-3ax^4+3ax^2-a+bx^2-b=a^2x^6+2abx^4+b^2x^2-1
a^2=a and -3a=2ab and 3a+b=b^2 and -a-b=-1. Since a!=0 a=1 and b=0 but it does not verify relations in the middle
In a general case P(X)=X*E where E is an even polynomial
=> (X^2-1)E(x^2-1)=x^2(E(X))^2-1
(X^2-1)E(x^2-1)=(x^2-1)(E(X))^2+(e(X))^2-1. So e(X) = 1*x^2m+sum(a2i * x^2i *(x^2-1)) a term that can factor out x^2-1 when dividing 1 and a part divisible by X^2-1 . This looks quite bad
Going back to the odd polynomial if X=I then P(-2)=(P(I))^2-1 which has th sum of alternating sign coefficients inside the paranthesis P(2)=(P(√3))^2-1 which is the negated value of P(2) but each coefficient has different multiplier. Overall (P(√3))^2-1=-(P(I))^2+1
(P(√3)+P(I))(P(√3)-P(I))=2 I guess I am not clever enough for this Olympiad problem. This will get a relationship between coefficients but it won't be pretty
Case p(X) even then P(0)=c
X=1 P(0)=(P(1))^2-1=c
X=0 P(-1)=P(1)=(P(0))^2-1
=> P(1)=c^2-1 and (P(1))^2=c+1
(C^2-1)^2=c+1
C^4-2c^2+1=c+1
C(c^3-2c-1)=0
c(c+1)(c^2-c-1)=0
We have a few cases here:
C=0 means p(1)=+1 or -1 the rank of polynomial is >0 so from Sybermath's observation the leading coefficient is +1
=> P(X)=x^2k because the sum of coefficients is also 1
There are infinite solutions actually. You can iterate the polynomial x^2 +1 to get more solutions. See math stackexchange question 271337 for a discussion on this exact problem.
@@gregoryknapen9133 Right!
There are infinitely many polynomials that satisfy the eq given.
P(x) = c where c = c^2 -1
P(x) = x
P(x) = x^2 - 1
P(x) = x^4 - 2x^2
...
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