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.  Reply With Quote

2.

3. Originally Posted by Fantini 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.  Reply With Quote

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

Cheers.  Reply With Quote

5. Originally Posted by Fantini 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.   Reply With Quote

6. 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.  Reply With Quote

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!  Reply With Quote

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