A problem of factoring a polynomial with a hint
$begingroup$
PT $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ can not be factored into two smaller polynomial $P(x)$ and $Q(x)$ with integer coefficients, where $a_{i}$'s are all different integers.
This problem can be solved by considering the root of equation $P(x)Q(x)-1=0$
This problem comes from Terry Tao's book Solving mathematical problem(page 47), in which he gives a hint as
if P(x) and Q(x) are such factors then what can you say about $P(x)-Q(x)$
How does one solve this problem this hint?
Edit:
This appears not to be true as pointed out by Darji and Eric. For interested readers, The actual problem can be found here, page 47 Excercise 3.7
polynomials contest-math irreducible-polynomials
asked 4 hours ago
zimbra314zimbra314
528212
$endgroup$
add a comment |
$begingroup$
PT $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ can not be factored into two smaller polynomial $P(x)$ and $Q(x)$ with integer coefficients, where $a_{i}$'s are all different integers.
This problem can be solved by considering the root of equation $P(x)Q(x)-1=0$
This problem comes from Terry Tao's book Solving mathematical problem(page 47), in which he gives a hint as
if P(x) and Q(x) are such factors then what can you say about $P(x)-Q(x)$
How does one solve this problem this hint?
Edit:
This appears not to be true as pointed out by Darji and Eric. For interested readers, The actual problem can be found here, page 47 Excercise 3.7
polynomials contest-math irreducible-polynomials
asked 4 hours ago
zimbra314zimbra314
528212
$endgroup$
1
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
3 hours ago
add a comment |
$begingroup$
PT $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ can not be factored into two smaller polynomial $P(x)$ and $Q(x)$ with integer coefficients, where $a_{i}$'s are all different integers.
This problem can be solved by considering the root of equation $P(x)Q(x)-1=0$
This problem comes from Terry Tao's book Solving mathematical problem(page 47), in which he gives a hint as
if P(x) and Q(x) are such factors then what can you say about $P(x)-Q(x)$
How does one solve this problem this hint?
Edit:
This appears not to be true as pointed out by Darji and Eric. For interested readers, The actual problem can be found here, page 47 Excercise 3.7
polynomials contest-math irreducible-polynomials
asked 4 hours ago
zimbra314zimbra314
528212
$endgroup$
PT $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ can not be factored into two smaller polynomial $P(x)$ and $Q(x)$ with integer coefficients, where $a_{i}$'s are all different integers.
This problem can be solved by considering the root of equation $P(x)Q(x)-1=0$
This problem comes from Terry Tao's book Solving mathematical problem(page 47), in which he gives a hint as
if P(x) and Q(x) are such factors then what can you say about $P(x)-Q(x)$
How does one solve this problem this hint?
Edit:
This appears not to be true as pointed out by Darji and Eric. For interested readers, The actual problem can be found here, page 47 Excercise 3.7
polynomials contest-math irreducible-polynomials
polynomials contest-math irreducible-polynomials
asked 4 hours ago
zimbra314zimbra314
528212
asked 4 hours ago
zimbra314zimbra314
528212
edited 18 mins ago
zimbra314
asked 4 hours ago
zimbra314zimbra314
528212
asked 4 hours ago
zimbra314zimbra314
528212
asked 4 hours ago
zimbra314zimbra314
528212
528212
1
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
3 hours ago
add a comment |
1
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
3 hours ago
1
1
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
3 hours ago
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.
But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.
The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.
answered 3 hours ago
Eric WofseyEric Wofsey
184k13213339
$endgroup$
$begingroup$
Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
$endgroup$
– zimbra314
21 mins ago
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
2 mins ago
add a comment |
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1 Answer
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$begingroup$
Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.
But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.
The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.
answered 3 hours ago
Eric WofseyEric Wofsey
184k13213339
$endgroup$
$begingroup$
Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
$endgroup$
– zimbra314
21 mins ago
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
2 mins ago
add a comment |
$begingroup$
Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.
But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.
The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.
answered 3 hours ago
Eric WofseyEric Wofsey
184k13213339
$endgroup$
$begingroup$
Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
$endgroup$
– zimbra314
21 mins ago
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
2 mins ago
add a comment |
$begingroup$
Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.
But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.
The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.
answered 3 hours ago
Eric WofseyEric Wofsey
184k13213339
$endgroup$
Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.
But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.
The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.
answered 3 hours ago
Eric WofseyEric Wofsey
184k13213339
edited 3 hours ago
answered 3 hours ago
Eric WofseyEric Wofsey
184k13213339
answered 3 hours ago
Eric WofseyEric Wofsey
184k13213339
answered 3 hours ago
Eric WofseyEric Wofsey
184k13213339
184k13213339
$begingroup$
Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
$endgroup$
– zimbra314
21 mins ago
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
2 mins ago
add a comment |
$begingroup$
Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
$endgroup$
– zimbra314
21 mins ago
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
2 mins ago
$begingroup$
Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
$endgroup$
– zimbra314
21 mins ago
$begingroup$
Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
$endgroup$
– zimbra314
21 mins ago
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
2 mins ago
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
2 mins ago
add a comment |
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$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
3 hours ago