Find all integral solutions of equation $$x^n+y^n+z^n=2016,$$ where $x,y,z,n -$ integers and $n\ge 2$
My work so far:
1) $n=2$ $$x^2+y^2+z^2=2016$$
I used wolframalpha n=2 and I received the answer to the problem (Number of integer solutions: 144)
2) $n=3$
I used wolframalpha n=3 and I not received the answer to the problem
How to do without wolframalpha?
On
Here an incomplete answer about the possibilities for $2016$. It is clear that the problem would become more difficult for higher numbers. $$\boxed {n=2}$$ It is known enough that an integer $n$ is a sum of three squares if and only if $n$ is not of the form $4^a(8b+7)$ so because $2016=4^2(8\cdot15+6)$ we can ensure that $x^2+y^2+z^2=2016$ has solutions. Calculation gives for integer non negatives $$(x,y,z)\in\{(4,8,44),(4,20,40),(12,24,36)\}$$ The other solutions are obtained trivially by permutations and change of signs so we get the $3\cdot8\cdot6=144$ solutions the OP said given by Wolphram. $$\boxed {n=3}$$ The actual state of the question has been well explained by Dietrich Burde above. $$\boxed {n=4}$$ We try to see this case in some detail. Note that $6^4=1296<2016<7^4=2401$ so we have to see the $4$-th powers $0,1,16,81,256,625,1296$ and see at the equations $$\begin{cases}y=\sqrt[4]{2016-z^4}\\y=\sqrt[4]{2015-z^4}\\ y=\sqrt[4]{2000-z^4}\\ y=\sqrt[4]{1935-z^4}\\ y=\sqrt[4]{1760-z^4}\\ y=\sqrt[4]{1391-z^4}\\ y=\sqrt[4]{720-z^4 }\end{cases}$$ where the numbers are $2016-x^4$ corresponding to $x=0,1,2,3,4,5,6$ respectively. There are no solutions (excepting some mistake in calculations). $$\boxed {n=6,8,…2k}$$ Since $3^6\lt 2016\lt 4^6$ and $2^8\lt 2016\lt 3^8$ and $2^{10}\lt 2016\lt 2^{11}$ it is easy to examine the possibilities (apparently no solutions). $$\boxed {n=2k+1}$$ It seems that this problem is similar to that in which $n = 3$ due to the appearance of negative integers.
On
An approach that can sometimes help to find solutions to this type of equation is to consider the prime factors of the given integer. In this case:
$$2016 = 2^5.3^2.7 = 2^5(2^6-1) = 2^5((2.2^5)-1)$$
Hence:
$$\mathbf{2016 = 4^5 + 4^5 + (-2)^5}$$
And also (finding a solution of $x^3+y^3+z^3=252$ by trial and error or from this list):
$$2016 = 2^3.252 = 2^3(7^3 - 3^3 - 4^3)$$
Hence:
$$\mathbf{2016 = 14^3 + (-6)^3 + (-8)^3}$$
The case $n=3$ has been studied intensively. Since $2016\equiv 0\bmod 9$ there should be solutions (probably even infinitely many) of $x^3+y^3+z^3=2016$, but many of them will be very large. For a reference here see this MO-question. For example, the smallest solution to $x^3+y^3+z^3=30$ is $(x,y,z)=(−283059965,−2218888517,2220422932)$, and for $x^3+y^3+z^3=33$ the status is already unknown.