The good thing about programming with Mathematica is that you can move from conceptual problems like the one described to getting a solution pretty easily. Since you're new to Mathematica, let's walk this through step by step:
1. Let $M$ be an $n\times n$ matrix
This one's easy:
ℳ[n_Integer] := Array[m, {n, n}]
ℳ[5] // MatrixForm

2. A function $m:\ \mathbb{N}\times\mathbb{N}\to \mathbb{Z}$ such that $m_{ij}=m_{ji}$
In other words, we don't care about the order $i,j$ or $j,i$ — we can achieve this simply by making m have the attribute Orderless.
SetAttributes[m, Orderless]
ℳ[5] // MatrixForm

Compare this output with the one above.
3. A function $m:\ \mathbb{N}\times\mathbb{N}\to \mathbb{Z}$ such that $m_{ij}=m_{i+2,j+2}$
Here, we can use RSolve to solve for the relation m[i,j] == m[i+2,j+2]:
RSolve[m[i, j] == m[2 + i, 2 + j], m[i, j], {i, j}]
(* {{m[i, j] -> C[1][-2 i + 2 j] + (-1)^i C[2][-2 i + 2 j]}} *)
What this solution says is that $m$ can be a function of the form
$$m: (i,j)\mapsto C_1(i,j) +(-1)^iC_2(i,j)$$
where the $C_i$ are constant functions (C denotes a constant in Mathematica).
Now putting all of these together, we can do:
Block[{m, p, q},
SetAttributes[m, Orderless];
m[i_, j_] := p + (-1)^i q;
ℳ[5]
] // MatrixForm

where p and q are constants you can choose.
nis specified one can use FindInstance to solve for the m_i,j. – Daniel Lichtblau Feb 26 at 20:41M. Is that not the case? (In particular, is yourMa numeric matrix?) If so then I'm not clear as to what is the question. – Daniel Lichtblau Feb 26 at 20:49