# Solving $n^5+n^4-3=x^2\pmod p$

Prove that for every odd prime number $p$ there is a natural number $n$ such that the equation $n^5+n^4-3=x^2\pmod p$ has no solutions.

So we have to understand that for each $p$ we can find $n$ such that the Legendre symbol $\left(\dfrac{n^5+n^4-3}{p}\right) = -1$. For $p=4k+3$ we can take $n=1$. How to deal with $4k+1$ case?

• Is x fixed, or does it vary with $p$? Commented Dec 6, 2014 at 6:51
• Firstly we fix $p$ then we have to find $n$ such that this eqaution with variable $x$ has no solutions. Commented Dec 6, 2014 at 6:52
• OK, comment deleted: I'd misread the Legendre symbol, as usual. Though I think "reduced" is a bit strong: it's just a rewording using the symbol notation, isn't it?
– HTFB
Commented Dec 9, 2014 at 17:02

The curve $x^2=n^5+n^4-3$ has genus $2$, and so the Hasse-Weil bound implies that the number of ordered pairs $(x,n)$ modulo $p$ satisfying the equation is between $p-4\sqrt p$ and $p+4\sqrt p$. There can be at most five values of $n$ for which the right-hand side is congruent to $0\pmod p$; for all other values of $n$, if $x$ satisfies the congruence then so does $-x$. Therefore the number of distinct $n\pmod p$ that can appear is at most $5 + \frac12(p+4\sqrt p-5)$. This quantity is less than $p$ as soon as $p\ge29$, and so there exists $n\pmod p$ such that the equation has no solutions. And the result can be checked by hand for $3\le p\le 23$.
• When $n=2$, we have $n^5+n^4-3=3^2\cdot 5$ so the equation has a solution modulo $p$ if and only if $\left(\frac{5}{p}\right)=1$. If $p=4k+1$, by quadratic reciprocity this is $\left(\frac{p}{5}\right) \equiv (4k+1)^{\frac{5-1}{2}}\equiv (k-1)^2$ modulo 5. This is $1$ if and only if $k \equiv 0$ or $2$ modulo $5$, which implies $p\equiv 1$ or $9$ modulo $20$, and the smallest such prime is $29$. Commented Dec 10, 2014 at 7:43