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Let $n,N$ be two natural numbers. Let $a=(a_1, ..., a_n),b=(b_1, ..., b_n)\in (\mathbb{Z}/N\mathbb{Z})^n$ with $\gcd(a_1, ..., a_n, N) =1$ and $\gcd(b_1, ..., b_n, N) =1$. Define $H_a = \{(x_1,...,x_n)\in(\mathbb{Z}/N\mathbb{Z})^n : a_1x_1+...+a_nx_n \equiv 0 \mod N\}$ and similarly define $H_b$. What is a necessary and sufficient condition for $H_a = H_b$ to hold? I suspect the condition is $(a_1, ..., a_n)= k\cdot(b_1, ..., b_n)$ for some $ k \in (\mathbb{Z}/N\mathbb{Z})^*$ since it's a sufficient condition and for $n=1$ I know it's also necessary. Can someone help me treat the $n \geq 2$ case?

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    Take $N=pqr$ for primes $p<q<r$. Then for every equation $Apqx+Bpry+Cqrz \equiv 0 (N)$, where $A,B,C$ are coefficients coprime to $N$, the set ${(x,y,z)\in \Bbb Z/N\Bbb Z: x \equiv 0 (r), y \equiv 0 (q), z \equiv 0 (p)}$ is its solution set (these congruences follow from the equation if we multiply the LHS by 2-products, and clearly they are solutions). – Amateur_Algebraist Dec 09 '23 at 19:18
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    @Amateur_Algebraist I see, thanks! So what will be a correct condition? –  Dec 09 '23 at 20:09
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    V. P. Elizarov considered such equations (and systems) in a 2002 paper (in Russian), I won't promise but I might go through it later. Statement 4.4 gives a formula for the general solution of a single inhomogeneous equation. – Amateur_Algebraist Dec 09 '23 at 21:26
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    This is an answer to the (first) question, not just a comment. Please post your answer as an answer. This brings extra visibility to the answer und puts the question off the unanswered list (otherwise it will be bumped up periodically). Also notice that comments are not indexed by the full text search, don't have any revision history, can only be edited for 5 minutes and allow only limited markup. Comments should only be used to clarify, not answer the problem. See How do comments work for more information. – Martin Brandenburg Dec 09 '23 at 21:29
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    @Amateur_Algebraist That would be helpful if you can. I'll try myself, even though I'm not speaking Russian unfortunately –  Dec 09 '23 at 22:27

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Let $R = \Bbb Z/N\Bbb Z$. I'll closely follow this 2002 paper by V. P. Elizarov.

Lemma 4.2. If $a, b \in R$ and $\gcd(a,b)=d$ in $\Bbb Z$, then there exists a matrix $Q \in GL_2(R)$ s.t. $(a,b)Q = (d,0)$.

Proof. If $a = 0$ or $b = 0$, take $Q$ to be the identity or an antidiagonal matrix with $1$'s. Suppose $a, b \not = 0$. Over $\Bbb Z$, express $d = a u_1 + b v_1$ by Bezout's identity. If $a = df$ and $b = dg$, then $1 = f u_1 + g v_1$ in $\Bbb Z$.

Furthermore, denote $u = u_1 \mod N$ and $v = v_1 \mod N$ ($u,v \in R$); then we have $d = au+bv$ in $R$, $1=fu+gv$ in $R$. Take $Q = (\begin{smallmatrix}u&g\\v&-f\end{smallmatrix})$ (it seems that the second column in his paper was upside down). Then $AQ = (au + bv, ag-bf) = (1,0)$; the determinant is $-uf-vg=-1$ which means $Q$ is invertible.

Statement 4.3. If $\gcd(a_1, \ldots, a_n) = d$, then there exists $V \in GL_n(R)$ s.t. $AV = (d, 0, \ldots, 0)$, where $A = (a_1,\ldots,a_n)$.

Proof. Start from $(a_{n-1}, a_n)$ and successively eliminate the nonzero elements using Lemma 4.2 (going backwards). Then take the product of these matrices, padded with blocks of identity transformations.

Special case of Statement 4.4. The set of solutions of $A (x_1, \ldots, x_n)^T = 0$ over $R$, where $\gcd(\{a_i\}, n) = 1$, is $V (0, *, \ldots, *)^T$ (the asterisks are arbitrary elements of $R$).

The proof is said to be "a direct check".

Denote $S = (0, *, \ldots, *)^T$. Then, for $A$ and $B$, we can find the corresponding matrices $V_A$ and $V_B$ and ask whether $V_A S = V_B S$, or equivalently, $V_B^{-1}V_A S = S$. Both sides have equal cardinalities, so we only need to check that $V_B^{-1} V_A e_k \in S$ for $k = 2, \ldots, n$ ($\{e_k\}$ is a "standard basis" of $S$, i.e. vectors of all zeros except one $1$).

Amateur_Algebraist
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