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Thread: Polynomials

  1. MHB Craftsman
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    #1
    I'm having trouble with the following question:

    Construct a polynomial $q(x) \neq 0$ with integer coefficients which has no rational roots but is such that for any prime $p$ we can solve the congruence $q(x) \equiv 0 \mod p$ in the integers.

    Any hints on how to even start the problem will be strongly appreciated. Thanks.

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  3. MHB Master
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    #2
    Quote Originally Posted by Fantini View Post
    I'm having trouble with the following question:

    Construct a polynomial $q(x) \neq 0$ with integer coefficients which has no rational roots but is such that for any prime $p$ we can solve the congruence $q(x) \equiv 0 \mod p$ in the integers.

    Any hints on how to even start the problem will be strongly appreciated. Thanks.
    Hi Fantini,

    Let me give you a hint. Consider the and see whether you can think of a polynomial satisfying the given conditions.

    Kind Regards,
    Sudharaka.

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    Thank you Sudharaka, but now I have found the solution. I will post it later.

    Cheers.

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    #4
    Quote Originally Posted by Fantini View Post
    Thank you Sudharaka, but now I have found the solution. I will post it later.

    Cheers.
    You are welcome. I was thinking about the polynomial \(q(x)=x^p-x\) so that the congruence \(q(x)\equiv 0(\mbox{mod } p)\) could be solved in the integers (by the Fermat's little theorem), but alas \(q\) has rational roots (examples are \(x=0,1\)). So to make it rational root less, we can do a little improvement. Take,

    \[q(x)=x^p-x+p\]

    By the the list of possible rational roots that this polynomial could have are \(\pm 1,\pm p\). Clearly, \(x= \pm 1, p\) do not satisfy the equation \(q(x)=0\). Let \(x=-p\) be a root of \(q(x)\). Then,

    \[q(-p)=0\Rightarrow (-p)^p+2p=0\]

    If \(p\) is even, \(p=2\) and we have \(p^p+2p=8=0\) which is contradictory. If \(p\) is odd,

    \[\Rightarrow -p^p+2p=0\Rightarrow p^{p-1}=2\]

    Since \(p\) is an odd prime, \(p\geq 3\). Hence the equation \(p^{p-1}=2\) cannot be satisfied. Therefore we see that there are no rational roots for,

    \[q(x)=x^p-x+p\]

    Interested to see your solution.

  6. MHB Oldtimer
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    #5
    A quick internet search comes up with the polynomial $p(x) = (x^2-2)(x^2-3)(x^2-6)$ (see , for example).

    The reason that works is that if 2 and 3 are both quadratic non-residues mod $p$ then their product 6 will be a quadratic residue. On the other hand, $p(x)$ clearly has no rational roots.

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    #6 Thread Author
    The solution is similar to what Opalg's search found. We considered the polynomial $q(x) = (x^2 +1)(x^2 +2)(x^2 -2)$ and, using elementary number theory arguments such as Legendre's symbol and quadratic residues, proved it satisfied the conditions of the statement.

    Thanks for the search link, Opalg!

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