Solving multivariate quadratic equations over the integers I am looking for a method (if it exists) to solve over the integers the following sum of squares equation:
$$ x_1^2 + x_2^2+x_3^2 + \cdots + x_n^2 = m,$$ with $m \in \mathbb{N}.$
Someone has any idea about books, articles dealing with this kind of problem?
Thanks in advance!
 A: If you just want sequential integers:
$\qquad m=(1^2+2^2+3^2+\dots,n^2)=\sum_{i=1}^n i^2=\dfrac{n(n+1)(2n+1)}{6}. \tag{1}$
For other sums of squares, you must know what the individual values or their limits, e.g.
$5^2+6^2+7^2+8^2\quad
=\sum_{i=1}^8 i^2-\sum_{i=1}^4 i^2\quad
\text{using the formula above} \tag{2}$
If you know how many square are identical, just multiply by that number those squares.
If you know sums that are discontinuous in spots, calculate the sequential parts separately as in equation
$\space (2)\space$ above.
For "sum" formulas up to $x^{10}$ look here.
A: If $m\in\mathbb{N}$, then $x_n\in\mathbb{N}$ meaning that any $x$-values will work. If $m$ is a perfect square, the are also infinite solutions because the Pythagorean theorem works for any number of dimensions.
For every Pythagorean triple, there is a Pythagorean quadruple where $C$ of the triple is replaced by $(A,B)$ of some other triple, and the pattern continues indefinitely. To explain, let's use a backwards example. In the triple
$(5,12,13),$ we can replace $5$, with $(3,4)$ because $3^2+4^2=5^2 \implies 3^2+4^2+12^2=13^2.$
To find triples in the forward direction, we need only consider that there is a side A for ever odd number greater than $1$. If we define Euclid's formula as
$ \quad A=m^2-k^2\quad B=2mk \quad C=m^2+k^2\quad$  we can solve the $C$-function for $k$ and find that any $m$-value that yields an integer $k$ provides the $(m,k)$ needed for a Pythagorean triple.
Here is how an $A=5$ triple is found to replace the $C$-value in $(3,4,5)$
\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\\
\text{for}\qquad  \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.
$$A=5\implies \lfloor\sqrt{5+1}\rfloor=2\le m \le \frac{5+1}{2} =3\\
\land
\quad m\in\{3\}\implies k \in\{2\} $$
$$F(3,2)=(5,12,13)$$
A similar process will find triples for $A=13$ and so on indefinitely to form n-tuples such as where
$$3^2+4^2+12^2+84^2+132^2+12324^2+6900^2+1428^2=14197^2\\
\text{ or }\quad 
15^2+8^2+144^2+404^2+93744^2=93745^2$$
