I have $NP$ equations in the form: $$y_{n,p} = x_{n} + b_{p},$$ with $n = 0, 1, ..., N$ and $p = 0, 1, ..., P$.
I know each $y_{n,p}$, but I don't know any $x_{n}$ or any $b_{p}$. I would like to determine each $x_{n}$ from the $y_{n,p}$. I don't need to know $b_{p}$. Since I have $NP$ linear equations for $N+P$ unknowns, I would expect a solution as long as $NP \geq N+P$ (which is verified as soon as $N\geq 2$ and $P\geq 2$).
For example, with $N=3$ and $P=2$, we have $$\left\{\begin{aligned} y_{1,1} &= x_{1} + b_{1} \\ y_{2,1} &= x_{2} + b_{1} \\ y_{3,1} &= x_{3} + b_{1} \\ y_{1,2} &= x_{1} + b_{2} \\ y_{2,2} &= x_{2} + b_{2} \\ y_{3,2} &= x_{3} + b_{2} \end{aligned} \right.$$
This is different than traditional systems of linear equations, since only additions are involved. It looks very simple, but I cannot find a way to solve it, and in fact I am not sure we can solve it. I can play with the equations to eliminate the $b_p$ (e.g. $y_{2,1} - y_{1,1} = x_{2} - x_{1}$) or the $x_n$ (e.g. $y_{1,2} - y_{1,1} = b_{2} - b_{1}$), but I don't see how to express each $x_n$ from $y_{n,p}$ only.
I think the problem as stated, does not have a solution.
I will use the $N=2$ and $P=3$ example to illustrate why
$$\begin{bmatrix}y_{1,1}\\ y_{1,2}\\ y_{1,3}\\ y_{2,1}\\ y_{2,2}\\ y_{2,3} \end{bmatrix}=\begin{bmatrix}1\\ 1\\ 1\\ & 1\\ & 1\\ & 1 \end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}+\begin{bmatrix}1\\ & 1\\ & & 1\\ 1\\ & 1\\ & & 1 \end{bmatrix}\begin{bmatrix}b_{1}\\ b_{2}\\ b_{3} \end{bmatrix}\tag{1}$$
I assign names to the above vectors and matrices such as the expression below is the same as the expression above
$$ \boldsymbol{y}={\rm G}\boldsymbol{x}+{\rm H}\boldsymbol{b} \tag{1} $$
where $\boldsymbol{x} = [x_1,x_2 \ldots x_N]$, $\boldsymbol{b} = [b_1,b_2,\ldots b_P]$, and $\boldsymbol{y} = [y_{1,1}\ldots, y_{N,P}]$
Note the projections below
$$\begin{gathered}{\rm G}^{\intercal}{\rm G}=\begin{bmatrix}3\\ & 3 \end{bmatrix}=P\;{\bf 1}\\ {\rm H}^{\intercal}{\rm H}=\begin{bmatrix}2\\ & 2\\ & & 2 \end{bmatrix}=M\;{\bf 1}\\ {\rm G}^{\intercal}{\rm H}=\begin{bmatrix}1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix} \end{gathered} \tag{2}$$
Here ${\bf 1}$ denotes an identity matrix.
Using the above projections we can eliminate one the two unknown vectors at a time to try to solve for the other one
$$\begin{gathered}{\rm G}^{\intercal}\boldsymbol{y}=P\boldsymbol{x}+{\rm G}^{\intercal}{\rm H}\boldsymbol{b}\\ \boldsymbol{x}=\tfrac{1}{P}{\rm G}^{\intercal}\left(\boldsymbol{y}-{\rm H}\boldsymbol{b}\right)\\ \left({\bf 1}-{\rm G}\tfrac{1}{P}{\rm G}^{\intercal}\right)\boldsymbol{y}=\underbrace{\left({\bf 1}-{\rm G}\tfrac{1}{P}{\rm G}^{\intercal}\right)}_{\text{singular}}{\rm H}\boldsymbol{b} \end{gathered} \tag{3}$$
and
$$ \begin{gathered}{\rm H}^{\intercal}\boldsymbol{y}={\rm H}^{\intercal}{\rm G}\boldsymbol{x}+N\boldsymbol{b}\\ \boldsymbol{b}=\tfrac{1}{N}{\rm H}^{\intercal}\left(\boldsymbol{y}-{\rm G}\boldsymbol{x}\right)\\ \left({\bf 1}-{\rm H}\tfrac{1}{N}{\rm H}^{\intercal}\right)\boldsymbol{y}=\underbrace{\left({\bf 1}-{\rm H}\tfrac{1}{N}{\rm H}^{\intercal}\right)}_{\text{singular}}{\rm G}\boldsymbol{x} \end{gathered} \tag{4}$$
Unfortunately, the singular coefficient matrix in front of the unknown vectors prevents us from solving the problem, regardless of the values for $N$ and $P$. Even the $N=P=2$ situation results in an underconstrained problem.