[proofplan] Fix an integer tuple $a\in\mathbb{Z}^n$ and compare the two conditions at that tuple. If every polynomial $F_j$ vanishes at $a$, then the sum of their squares vanishes immediately. Conversely, if the sum of squares vanishes, then each summand is a nonnegative integer, and a finite sum of nonnegative integers can be zero only when every summand is zero. [/proofplan]

[step:Evaluate the sum of squares at a fixed integer tuple] Fix $a=(a_1,\ldots,a_n)\in\mathbb{Z}^n$. For each $j\in\{1,\ldots,m\}$, define the integer \begin{align*} b_j:=F_j(a_1,\ldots,a_n). \end{align*} Since $F_j\in\mathbb{Z}[x_1,\ldots,x_n]$ and $a_i\in\mathbb{Z}$ for every $i\in\{1,\ldots,n\}$, polynomial evaluation gives $b_j\in\mathbb{Z}$. By the definition of $G$, \begin{align*} G(a_1,\ldots,a_n)=\sum_{j=1}^{m}b_j^2. \end{align*} [/step]

[step:Show that a solution of the system solves the single equation] Assume that $F_j(a_1,\ldots,a_n)=0$ for every $j\in\{1,\ldots,m\}$. Then $b_j=0$ for every $j$, so $b_j^2=0$ for every $j$. Therefore \begin{align*} G(a_1,\ldots,a_n)=\sum_{j=1}^{m}b_j^2=0. \end{align*} Thus every integral solution of the system is an integral solution of the equation $G=0$. [/step]

[step:Show that a solution of the single equation solves the system]Assume that \begin{align*} G(a_1,\ldots,a_n)=0. \end{align*} Using the notation $b_j=F_j(a_1,\ldots,a_n)$ from the first step, this means \begin{align*} \sum_{j=1}^{m}b_j^2=0. \end{align*} For every $j\in\{1,\ldots,m\}$, the square $b_j^2$ is a nonnegative integer. Suppose, toward a contradiction, that $b_k\ne 0$ for some $k\in\{1,\ldots,m\}$. Since $b_k\in\mathbb{Z}$, this implies $|b_k|\ge 1$, hence \begin{align*} b_k^2\ge 1. \end{align*} All other summands $b_j^2$ are nonnegative, so \begin{align*} \sum_{j=1}^{m}b_j^2\ge b_k^2\ge 1, \end{align*} contradicting $\sum_{j=1}^{m}b_j^2=0$. Hence $b_j=0$ for every $j\in\{1,\ldots,m\}$, which is exactly \begin{align*} F_j(a_1,\ldots,a_n)=0 \text{ for every } j\in\{1,\ldots,m\}. \end{align*}[/step]

[guided]We assume that the single equation holds at the fixed integer tuple $a$, namely \begin{align*} G(a_1,\ldots,a_n)=0. \end{align*} By the definition of $G$ and by the notation $b_j:=F_j(a_1,\ldots,a_n)$, this is the same as \begin{align*} \sum_{j=1}^{m}b_j^2=0. \end{align*} The key point is that the summands are not arbitrary integers; they are squares of integers. For each $j\in\{1,\ldots,m\}$, we have $b_j\in\mathbb{Z}$, so $b_j^2\in\mathbb{Z}$ and $b_j^2\ge 0$. Thus the displayed sum is a finite sum of nonnegative integers. We now prove that no nonzero $b_j$ can occur. Suppose that there is some index $k\in\{1,\ldots,m\}$ such that $b_k\ne 0$. Since $b_k$ is an integer, the condition $b_k\ne 0$ implies $|b_k|\ge 1$. Squaring gives \begin{align*} b_k^2\ge 1. \end{align*} Every other term $b_j^2$ is nonnegative, so the full sum satisfies \begin{align*} \sum_{j=1}^{m}b_j^2\ge b_k^2\ge 1. \end{align*} This contradicts the assumed equality $\sum_{j=1}^{m}b_j^2=0$. Therefore no such index $k$ exists, and hence $b_j=0$ for every $j\in\{1,\ldots,m\}$. Substituting back the definition $b_j:=F_j(a_1,\ldots,a_n)$, we obtain \begin{align*} F_j(a_1,\ldots,a_n)=0 \text{ for every } j\in\{1,\ldots,m\}. \end{align*} Thus any integral solution of the single equation $G=0$ is an integral solution of the original system.[/guided]

[step:Identify the two integral solution sets] We have shown that, for the fixed but arbitrary tuple $a\in\mathbb{Z}^n$, the condition \begin{align*} F_j(a_1,\ldots,a_n)=0 \text{ for every } j\in\{1,\ldots,m\} \end{align*} holds if and only if \begin{align*} G(a_1,\ldots,a_n)=0. \end{align*} Since $a\in\mathbb{Z}^n$ was arbitrary, the two conditions define the same subset of $\mathbb{Z}^n$. Therefore the integral solution set of the system $F_1=\cdots=F_m=0$ is equal to the integral solution set of the single equation $G=0$. [/step]