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For $k$ is a fixed number such that $-n^3 \leq k \leq 2n^3$. How many solutions to the equation $x^3 - y^3 + z^3 = k$ with $1 \leq x, y, z \leq n$?

One of the trivial bounds is $n^{1 + \epsilon}$ since we know that the number of divisors of $n$ is $n^{\epsilon}$. It is likely the trivial bounds can be achieved?

Now, we are interested in how many $k$ such that $x^3 - y^3 + z^3 = k$ has $n^{1 + \epsilon}$? Also, can we have a better bound for the number of solutions for the remaining $k$?

If we can consider a similar question for $x^d - y^d + z^d = k$, we have a uniform bound for the number of solutions by Heath-Brown. However, the Heath-Brown result is only non-trivial when $d$ is big enough.

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    $\begingroup$ How is the number of divisors of $n$ related to the number of solutions to the equation? $\endgroup$
    – Wojowu
    Commented Oct 19, 2023 at 21:26
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    $\begingroup$ @Wojowu: There are $O(n)$ choices for $z$. Fix any such choice with $z \neq k^{1/3}$, and note that $(x-y)(x^2+y^2+xy) = k - z^3$. The right hand side is $\ll n^3$, and hence has $\ll n^\epsilon$ divisors. For any such divisor, there are only $O(1)$ choices for both $x$ and $y$, and hence only $\ll n^{1+\epsilon}$ choices for solutions $(x,y,z)$ with $z \neq k^{1/3}$. The case $z = k^{1/3}$ remains, but there are only $O(n)$ such points as then $x = y, z = k^{1/3}$. $\endgroup$
    – Anurag Sahay
    Commented Oct 20, 2023 at 6:18
  • $\begingroup$ @TheNguyen: What type of answer to your question would you be satisfied with? $\endgroup$
    – Anurag Sahay
    Commented Oct 20, 2023 at 14:30

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