
MHB Craftsman
#1
September 27th, 2012,
13:32
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.

September 27th, 2012 13:32
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MHB Master
#2
October 29th, 2012,
10:42
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.

MHB Craftsman
#3
October 29th, 2012,
12:11
Thread Author
Thank you Sudharaka, but now I have found the solution. I will post it later.
Cheers.

MHB Master
#4
October 30th, 2012,
05:36
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 \(q(x)=x^px\) 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,
\[q(x)=x^px+p\]
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,
\[\Rightarrow p^p+2p=0\Rightarrow p^{p1}=2\]
Since \(p\) is an odd prime, \(p\geq 3\). Hence the equation \(p^{p1}=2\) cannot be satisfied. Therefore we see that there are no rational roots for,
\[q(x)=x^px+p\]
Interested to see your solution.

MHB Oldtimer
#5
October 30th, 2012,
13:52
A quick internet search comes up with the polynomial $p(x) = (x^22)(x^23)(x^26)$ (see , for example).
The reason that works is that if 2 and 3 are both quadratic nonresidues mod $p$ then their product 6 will be a quadratic residue. On the other hand, $p(x)$ clearly has no rational roots.

MHB Craftsman
#6
October 30th, 2012,
19:18
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!