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Bézout's theorem states that a linear diophantine equation $ax+by=c$ has integer solutions if and only if $\gcd(a,b)|c$. Then is there a theorem regarding the existence of a unique integer solution for a system of two linear equations $ax+by=c$ and $px+qy=m$, where $a,b,c,p,q,m$ are positive integers?

Additionally, can the theorem be generalized to the condition where there are more than two unknowns?

Kevin Xu
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    Consider the system over $\Bbb Q$. If the determinant of the matrix with $a,b,p,q$ is nonzero, we have a unique rational solution. So if there is an integer solution, it must be already unique. – Dietrich Burde Mar 21 '23 at 20:16
  • Given that you posted a question about matrix determinant ... Think about how leading zeros in the reduced echelon form of a matrix gives us restrictions on a unique solution. – Calvin Lin Mar 21 '23 at 20:17
  • Adding the link Calvin is mentioning here. – Dietrich Burde Mar 21 '23 at 20:18
  • I have already got this part regarding the uniqueness, but what I am focusing more on is the integer criteria – Kevin Xu Mar 21 '23 at 20:22
  • The integer criteria are simple. If the gcd divides the RHS in both cases, each equation has infinitely many integer solutions, and we can parametrize them, and then take both simultaneously. Then we obtain the ones for both equations. – Dietrich Burde Mar 21 '23 at 20:31
  • However, not all linear systems with all the equations satisfying the Bezout identity have an integer solution, e.g. $x+2y=3, 3x+4y=8$. That is why I am finding a determination method based on the parameters $a, b, c, p, q, m$ that can inform me whether a linear system has an integer solution or not – Kevin Xu Mar 21 '23 at 20:36
  • This might be of (tangential) interest. – joriki Mar 21 '23 at 20:47
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2 Answers2

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For two equations the problem actually becomes very simple as the qeustion is equivalent to whether or not there exists a vector $\begin{pmatrix}x\\y\end{pmatrix}\in\mathbb{Z}^2$ with $$\begin{pmatrix}a&b\\p&q\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}c\\m\end{pmatrix}.$$

If the determinant of $A=\begin{pmatrix}a&b\\p&q\end{pmatrix}$ is non-zero, we have a unique real solution, which we can easily calculate via calculating $A^{-1}$. If this solution is already integer, we have a unique integer solution, else we dont.

If the determinant of $A$ is zero, there are two more cases to differentiate. Either $a=b=p=q=0$, in which case $m=c=0$ as well and there are infinitely many integer solutions.

Or the determinant of $A$ is zero and at least one of $a,b,p,q$ is non-zero. In that case the kernel of $A$ is $1$-dimensional and the equations therefor admit either no or infinitely many real solutions. If there are infinitely many real solutions and $s$ is such a solution, then the set of all real solutions can be described by $$\left\lbrace v\in\mathbb{R}^2\ \Big|\ v=s+w\text{, where }w\in\text{ker}(A)\right\rbrace.$$ Since the kernel is 1-dimensional, it is generated by a single vector $k\in\mathbb{R}^2$, thus in the last case the set of equations admits an integer solution if and only if there exists a $\lambda\in\mathbb{R}$ with $$s+\lambda \cdot k\in\mathbb{Z}^2,$$ and it is unique if and only if there is no $\mu\in\mathbb{R}$ with $\mu\cdot k\in\mathbb{Z}^2$. Both of those equations are easy to solve.

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A necessary condition is: If we use Hermite normal form as mentioned in comments by @BillDubuque, we can assume $Ax = b$ with $A$ an upper triangular integer matrix.

A sufficient condition for the solution to exist: $$a_{ii} \ \ | \ \ gcd(a_{i,i+1},....,a_{i,n},b_i)$$.

A necessary condition for solution to exist: $$gcd(a_{ii},a_{i,i+1},....,a_{i,n}) \ \ | \ \ b_i $$.

If $a_{ii} = gcd(a_{i,i+1},....,a_{i,n})$ then the above necessary and sufficient condition become the same i.e., the solution exists iff $a_{ii} \ | \ b_i$ in this specific case. Need a necessary and sufficient condition in general though.

Good Luck !

Balaji sb
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