Let $V={\mathbb Q}^{\mathbb N}$ be the space of all functions ${\mathbb N}\to {\mathbb Q}$ where ${\mathbb N}=\lbrace 1,2,3,\ldots \rbrace$ is the set of all positive integers.
Let $W$ be the subspace of $W$ defined by
$$ W=\lbrace f\in V \ | \ f(a)+f(b)=f(c)+f(d) \ \textrm{if} \ a^2+b^2=c^2+d^2, \ a,b,c,d \in {\mathbb N}\rbrace $$
My question : What is the dimension of $W$ ?
What I did : I have shown that $\dim(W) \leq 6$ ; in fact, any $f(n)(n\in{\mathbb N})$ is a linear combination of $f(1),f(2),\ldots,f(6)$. Indeed, the identities $1^2+7^2=5^2+5^2$, $1^2+8^2=4^2+7^2$, $2^2+9^2=6^2+7^2$, $3^2+11^2=7^2+9^2$, $5^2+10^2=2^2+11^2$, yield succesively that this property is true for $7,8,9,11$, and $10$. Next, it is a simple enough arithmetical lemma that for $n\geq 12$, there are $a,b,c$ with $1\leq a,b,c \lt n$ such that $n^2+a^2=b^2+c^2$ (this is shown in different ways in different answers of an older MSE question ), so that we are then done.
Numerical experiments suggest that the dimension is exactly $6$.
UPDATE 10/04/2020 : The dimension of $W$ is at least four because $W$ contains the constants, the square function, the characteristic function of the numbers divisible by $4$ ($f(n)=1$ if $n$ is divisible by 4, and $f(n)=0$ otherwise), and the characteristic function of the odd numbers.
Here's what the beginning of the sequence corresponding to $f(1)=-1$, $f(k)=0$ for $2\leq k \leq 6$ looks like :
$-1, 0, 0, 0, 0, 0, 1, 2, 1, 2, 2, 4, 4, 4, 5, 6, 7, 8, 8, 10, 10, 12, 13, 14, 15, 16, 18, 20, 20, 22, 23, 26, 27, 28, 30, 32, 34, 36, 37, 40, 41, 44, 46, 48, 50, 52, 55, 58, 59, 62, 64, 68, 70, 72, 75, 78, 81, 84, 86, 90, 92, 96, 99, 102, 105, 108, 112, 116$
The sequence seems to be nondecreasing (from the ninth term on) and nonnegative (except for the first term).
The two missing functions are the characteristic function of the multiples of $3$, and the Legendre symbol $n\mapsto\left(\frac{n}{5}\right)$. To check that these functions satisfy the given property, observe that the function value depends only on the input mod $3$ (resp. mod $5$), so it is enough to check the condition for $1\leq a,b\leq 3$ (resp. $1\leq a,b\leq 5$), a finite computation. Combined with the set of four functions your found, this gives six linearly independent solutions, which must be a basis.
For your example $f(1)=-1$, $f(k)=0$ for $2\leq k\leq 6$, I solved a linear system to find $$ f(n)=-\frac{1}{8}\chi_{2\mathbb{Z}+1}(n) +\frac{1}{2}\chi_{4\mathbb{Z}}(n)-\frac{1}{2}+\frac{1}{40}n^2-\frac{2}{5}\left(\frac{n}{5}\right). $$