In the course of some research, I've run into the following problem. Define $$ \phi = |n|^{c} - |n_1|^{c} + |n_2|^{c} - |n-n_1+n_2|^c. $$ I'm interested in $c \in [2,4]$. The $n, n_1, n_2, n_3$, are integers.
Question: For fixed $\phi \in \mathbb{R}$ and $n \in \mathbb{Z}$, can we bound the number of integer solutions $n_1, n_2, n_3$?
If $c = 2$, then the equation is $$ \phi = 2(n_1-n_2)(n-n_1).$$ In this case, $\phi \in \mathbb{Z}$, and the number of its divisors is bounded by $C|\phi|^\epsilon$ for any $\epsilon >0$ (see, e.g. https://terrytao.wordpress.com/2008/09/23/the-divisor-bound/). So in this case, if we fix $\phi$ and $n$, then there are order $|\phi|^{\epsilon}$ possibilities for $n_1$ and $n_2$.
If $c = 4$, the same idea works. We have $$ \phi = (n_1 - n_2)(n-n_1)\Bigl(n^2 +n_1^2 + n_2^2 + (n-n_1+n_2)^2 + 2(n+n_2)^2\Bigr),$$ and again, if we fix $\phi$ and $n$, the number of possibilities for $n_1$ and $n_2$ is bounded by $|\phi|^\epsilon$.
If $c = 3$, matters get a bit messier because of the absolute values, but it seems the same approach works, if we divide into different cases depending on the signs of $n$, $n_1$, $n_2$, and $n-n_1+n_2$.
Main Question: Can we say anything for $c \in (2,3) \cup (3,4)$? Here $\phi$ need not be integral, so divisor counting is out. Any ideas on other approaches? It seems reasonable to expect a similar bound, but may be very difficult to prove. Insight or counterexamples would be appreciated.
NB: One still has a bound $$ |\phi| \geq C|n_1 - n_2||n-n_1| \max\{ |n|, |n_1|, |n_2| \}^{c-2}. $$
For almost all real values of $c$, the values $n^c$ for nonzero integer $c$ will be linearly independent over the rationals, which drastically reduces the possible number of solutions.