3
$\begingroup$

For any prime $p>5$, one of the numbers $$1^2+1=2,\ \ 2^2+1=5,\ \ 3^2+1=10=2\times5$$ is a quadratic residue modulo $p$. In 2014 I conjectured that each prime $p$ has a primitive root $g<p$ of the form $k^2+1\ (k\in\mathbb Z)$ (cf. http://oeis.org/A239957); this is still open.

By a result of Fermat, the equation $x^4+y^4=z^2$ has no positive integer solution.

In view of the above, here I ask the following question.

Question 1. Whether for each prime $p>541$ there is a number of the form $x^4+y^4$ (with $x,y\in\mathbb Z)$ which is not only smaller than $p$ but also a quadratic residue modulo $p$?

Actually, I even conjecture that for any prime $p>541$ with $p\not=941$ there is a prime $q<p$ of the form $x^4+y^4\ (x,y\in\mathbb Z)$ with $\left(\frac q p\right)=1$, where $(-)$ is the Legendre symbol. Of course, it is not yet proven that there are infinitely many primes of the form $x^4+y^4$.

The following question is similar to Question 1.

Question 2. Whether for each odd prime $p\not\in\{7,17,47,103\}$ there is a number $q<p$ of the form $x^4+y^4\ (x,y\in\mathbb Z)$ with $\left(\frac qp\right)=-1$?

In 2001 Heath-Brown [Acta Math. 186 (2001), 1-84] proved that there are infinitely many primes of the form $x^3+2y^3$ with $x,y\in\mathbb N=\{0,1,2,\ldots\})$. Motivated by this, here I pose the following question.

Question 3. Whether for each odd prime $p$ there is a prime $q<p$ with $\left(\frac qp\right)=-1$ such that $q=x^3+2y^3$ for some $x,y\in\mathbb N$ with $y+1$ prime?

I have checked Question 3 for all odd primes $p<2\times10^9$; see http://oeis.org/A344173 for related data. For example, the prime $q=3^3+2(3-1)^3=43$ is a quadratic nonresidue modulo the prime $p=457$.

Your comments are welcome!

$\endgroup$
1
  • $\begingroup$ Similarly, as $x^3+y^3=z^3$ has no positive integer solution, I conjecture that for any prime $p>37$ with $p\equiv1\pmod3$ there is a number $x^3+y^3\ (x,y\in\{1,2,3,\ldots\})$ which is smaller than $p$ and also a cubic residue modulo $p$. $\endgroup$ May 11 at 11:22
2
$\begingroup$

I have just found an answer to Question 1. Observe that $$1^4+2^4=17,\ 5^4+6^4 = 1921=17\times113,\ \mbox{and}\ 17\times1921=32657=8^4+13^4.$$ For any odd prime $p\not=17,113$, one of the three numbers $17,\ 1921,\ 32657$ is a quadratic residue modulo $p$. So Question 1 has a positive answer for $p>32657$. For primes $p$ with $541<p<32657$, we can use a computer to make a check.

This method does not work for Questions 2 and 3.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.