# Does each prime $p>541$ have a quadratic residue $x^4+y^4<p$?

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 $$g of the form $$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 $$q of the form $$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 $$q of the form $$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 $$q with $$\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 $$p=457$$.

• 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 $p$. May 11 at 11:22
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, we can use a computer to make a check.