Proof verification for polynomial solutions to $xP(x-1)=(x-2)P(x)$
$begingroup$
Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.
Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$
Here is my solution, is it right?
Replace $x$ with $0$ and $1$ and clearly, they equal $0$.
Suppose there is another positive root $rneq 1$.
Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.
Thus if there exists another positive root $rneq 1$, then we have an infinite number of roots which means $P(x)=0$.
Now suppose there is another negative root $c$.
Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.
Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.
So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.
number-theory proof-verification polynomials roots functional-equations
edited Dec 31 '18 at 12:45
Maria Mazur
50.7k1362126
asked Dec 31 '18 at 12:28
user627514user627514
393
$endgroup$
add a comment |
$begingroup$
Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.
Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$
Here is my solution, is it right?
Replace $x$ with $0$ and $1$ and clearly, they equal $0$.
Suppose there is another positive root $rneq 1$.
Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.
Thus if there exists another positive root $rneq 1$, then we have an infinite number of roots which means $P(x)=0$.
Now suppose there is another negative root $c$.
Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.
Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.
So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.
number-theory proof-verification polynomials roots functional-equations
edited Dec 31 '18 at 12:45
Maria Mazur
50.7k1362126
asked Dec 31 '18 at 12:28
user627514user627514
393
$endgroup$
$begingroup$
Why is $P(r+1)=0$?
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:36
$begingroup$
It is better if you go by carat foot steps...
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:37
2
$begingroup$
What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 12:59
add a comment |
$begingroup$
Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.
Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$
Here is my solution, is it right?
Replace $x$ with $0$ and $1$ and clearly, they equal $0$.
Suppose there is another positive root $rneq 1$.
Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.
Thus if there exists another positive root $rneq 1$, then we have an infinite number of roots which means $P(x)=0$.
Now suppose there is another negative root $c$.
Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.
Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.
So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.
number-theory proof-verification polynomials roots functional-equations
edited Dec 31 '18 at 12:45
Maria Mazur
50.7k1362126
asked Dec 31 '18 at 12:28
user627514user627514
393
$endgroup$
Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.
Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$
Here is my solution, is it right?
Replace $x$ with $0$ and $1$ and clearly, they equal $0$.
Suppose there is another positive root $rneq 1$.
Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.
Thus if there exists another positive root $rneq 1$, then we have an infinite number of roots which means $P(x)=0$.
Now suppose there is another negative root $c$.
Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.
Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.
So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.
number-theory proof-verification polynomials roots functional-equations
number-theory proof-verification polynomials roots functional-equations
edited Dec 31 '18 at 12:45
Maria Mazur
50.7k1362126
asked Dec 31 '18 at 12:28
user627514user627514
393
edited Dec 31 '18 at 12:45
Maria Mazur
50.7k1362126
asked Dec 31 '18 at 12:28
user627514user627514
393
edited Dec 31 '18 at 12:45
Maria Mazur
50.7k1362126
edited Dec 31 '18 at 12:45
Maria Mazur
50.7k1362126
edited Dec 31 '18 at 12:45
Maria Mazur
50.7k1362126
50.7k1362126
asked Dec 31 '18 at 12:28
user627514user627514
393
asked Dec 31 '18 at 12:28
user627514user627514
393
asked Dec 31 '18 at 12:28
user627514user627514
393
393
$begingroup$
Why is $P(r+1)=0$?
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:36
$begingroup$
It is better if you go by carat foot steps...
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:37
2
$begingroup$
What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 12:59
add a comment |
$begingroup$
Why is $P(r+1)=0$?
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:36
$begingroup$
It is better if you go by carat foot steps...
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:37
2
$begingroup$
What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 12:59
$begingroup$
Why is $P(r+1)=0$?
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:36
$begingroup$
Why is $P(r+1)=0$?
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:36
$begingroup$
It is better if you go by carat foot steps...
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:37
$begingroup$
It is better if you go by carat foot steps...
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:37
2
2
$begingroup$
What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 12:59
$begingroup$
What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 12:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).
So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.
answered Dec 31 '18 at 12:44
Maria MazurMaria Mazur
50.7k1362126
$endgroup$
$begingroup$
Was about to upvote when I realsied that you're not actually answering the question.
$endgroup$
– Git Gud
Dec 31 '18 at 17:29
$begingroup$
Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
$endgroup$
– Maria Mazur
Dec 31 '18 at 18:03
add a comment |
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1 Answer
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$begingroup$
If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).
So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.
answered Dec 31 '18 at 12:44
Maria MazurMaria Mazur
50.7k1362126
$endgroup$
$begingroup$
Was about to upvote when I realsied that you're not actually answering the question.
$endgroup$
– Git Gud
Dec 31 '18 at 17:29
$begingroup$
Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
$endgroup$
– Maria Mazur
Dec 31 '18 at 18:03
add a comment |
$begingroup$
If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).
So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.
answered Dec 31 '18 at 12:44
Maria MazurMaria Mazur
50.7k1362126
$endgroup$
$begingroup$
Was about to upvote when I realsied that you're not actually answering the question.
$endgroup$
– Git Gud
Dec 31 '18 at 17:29
$begingroup$
Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
$endgroup$
– Maria Mazur
Dec 31 '18 at 18:03
add a comment |
$begingroup$
If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).
So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.
answered Dec 31 '18 at 12:44
Maria MazurMaria Mazur
50.7k1362126
$endgroup$
If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).
So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.
answered Dec 31 '18 at 12:44
Maria MazurMaria Mazur
50.7k1362126
edited Dec 31 '18 at 13:03
answered Dec 31 '18 at 12:44
Maria MazurMaria Mazur
50.7k1362126
answered Dec 31 '18 at 12:44
Maria MazurMaria Mazur
50.7k1362126
answered Dec 31 '18 at 12:44
Maria MazurMaria Mazur
50.7k1362126
50.7k1362126
$begingroup$
Was about to upvote when I realsied that you're not actually answering the question.
$endgroup$
– Git Gud
Dec 31 '18 at 17:29
$begingroup$
Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
$endgroup$
– Maria Mazur
Dec 31 '18 at 18:03
add a comment |
$begingroup$
Was about to upvote when I realsied that you're not actually answering the question.
$endgroup$
– Git Gud
Dec 31 '18 at 17:29
$begingroup$
Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
$endgroup$
– Maria Mazur
Dec 31 '18 at 18:03
$begingroup$
Was about to upvote when I realsied that you're not actually answering the question.
$endgroup$
– Git Gud
Dec 31 '18 at 17:29
$begingroup$
Was about to upvote when I realsied that you're not actually answering the question.
$endgroup$
– Git Gud
Dec 31 '18 at 17:29
$begingroup$
Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
$endgroup$
– Maria Mazur
Dec 31 '18 at 18:03
$begingroup$
Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
$endgroup$
– Maria Mazur
Dec 31 '18 at 18:03
add a comment |
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$begingroup$
Why is $P(r+1)=0$?
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:36
$begingroup$
It is better if you go by carat foot steps...
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:37
2
$begingroup$
What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 12:59