A problem of factoring a polynomial with a hint












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
    $endgroup$
    – darij grinberg
    3 hours ago
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
    $endgroup$
    – darij grinberg
    3 hours ago














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 mins ago







zimbra314

















asked 3 hours ago









zimbra314zimbra314

523212




523212








  • 1




    $begingroup$
    What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
    $endgroup$
    – darij grinberg
    3 hours ago














  • 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










1 Answer
1






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    6 mins ago











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

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4












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    6 mins ago
















4












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    6 mins ago














4












4








4





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























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
    6 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
    6 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
6 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
6 mins ago


















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