I have a system of $4$ equations in $4$ variables:

\begin{align} x_1 + y_1 &= m\\ x_2 - y_1 &= n\\ x_1 - y_2& = o\\ x_2 + y_2 &= p\end{align}

$x_1, y_1, x_2, y_2$ are integer points on co-ordinate system (we need only positive points in the solution).

I want to have only the positive integer value for all the variables $(x_1, x_2, y_1, y_2)$.

Let's suppose \begin{align}m &= 3\\ n &= 10^9 - 1\\ o &= 10^9 - 3\\ p &= 2 \times 10^9 - 7\end{align}

So, the above equation get satisfied for the values:

\begin{align}x_1 &= -3\tag{here $x_1$ is negative}\\ x_2 &= 7\\ y_1 &= 6 \\ y_2 &= 0\end{align}

Whereas the following set of values also satisfies the equation:

\begin{align}x_1 &= 1\\ x_2 &= 3\\ y_1 &= 2\\ y_2 &= 4\tag{here none is negative}\end{align}

I just want to find out the solution which has non-negative integer values for all 4 variables (having $0$ in the solution set is fine, just avoid negative values) when $m, n, o$ and $p$ can be any given constant.

Conditions: \begin{align}0 \leq x_1 \leq x_2 \leq 10^9\\ 0 \leq y_1 \leq y_2 \leq 10^9\end{align}

  • $\begingroup$ I've edited your question to use MathJax instead of code for the equations. The above comment gives lots of help on this $\endgroup$
    – lioness99a
    Commented Feb 8, 2019 at 9:40

3 Answers 3


If you subtract eq. $3$ from eq. $1$, you get $$ y_1 + y_2 = m -o $$ If you subtract eq. $2$ from eq. $4$, you get $$ y_1 + y_2 = p-n $$ and therefore $m-o = p-n$. Similarly, of you add eq. 1 and 2, you get $$ x_1 + x_2 = m+n $$ and if you add eq. 3 and 4, you get $$ x_1 + x_2 = p+o $$ And therefore $m+n=p+o$. Your condition was that all the $x$'s and $y$'s have to be non-negative, so $$ m+n=o+p \geq 0 \qquad \text{and} \qquad m-o = p-n \geq 0 $$ Subtracting these from each other leads us to $m - p \geq 0$. The second equation then tells us $$ m-p = o-n \geq 0 \qquad \Rightarrow \qquad o\geq n $$ So now we have the conditions $$ m \geq p \qquad \text{and} \qquad o \geq n $$ in order for the $x$'s and $y$'s to be positive. Therefore, if these conditions are fulfulled, it's possible to choose values for the $x$'s and $y$'s so that they are positive.

  • $\begingroup$ That's fine. But the main thing is how to find the solution to the equations? $\endgroup$
    – Aman Gupta
    Commented Feb 8, 2019 at 10:02

Let's start with the linear equation system, forgetting the additional constraints for the moment.

Subtracting the third from the first equation gives


while subtracting the second from the fourth gives


So unless $m-o=p-n$ or equivalently $m+n=o+p$, your system doesn't have any solution.

So your equations are dependent on each other, and if we assume $m+n=o+p$ we can just remove one equation to get an equivalent system. I decided to remove the third equation, as that gets rid of the constant $o$ that is sometimes hard to distinguish from the number $0$.

$$\begin{array}{} x_1+y_1 & = & m \\ x_2-y_1 & = & n \\ x_2+y_2 & = & p \\ \end{array}$$

3 equations for 4 variables leaves one 'free choice' of a variable usually (unless there is even more dependence among the equations). So let's use $x_1$ as that variable.

The first equation immediately leads to


then the second equation leads to


and finally the third equation to


If you fear that having removed the 3rd original equation has been an error, you can see that it still holds:


taking into account the necessary condition $m+n=o+p$.

So back to the given constraints:

$0 \le x_1$ just becomes equivalently $x_1 \ge 0$.

$x_1 \le x_2$ becomes $x_1 \le n+m-x_1$, which is equivalent to $x_1 \le \frac{n+m}2$.

$x_2 \le 10^9$ becomes $x_1 \ge n+m - 10^9$.

Similiarly, $y_1 \ge 0$ becomes $x_1 \le m$, $y_1 \le y_2$ becomes $x_1 \ge m + \frac{n-p}2$ and $y_2 \le 10^9$ becomes $x_1 \le 10^9+n+m-p$.

That means all of your conditions are equivalent to

$$x_1 \ge \max\left\{0,n+m-10^9,m + \frac{n-p}2\right\}$$


$$x_1 \le \min\left\{\frac{n+m}2, m, 10^9+n+m-p\right\}.$$

So, always taking into account the necessary condition $m+n=o+p$, you can choose $x_1$ acording to those 2 inequalities, calculate $x_2,y_1,y_2$ according to the equations I gave above and get a solution that fulfills your other conditions as well.

  • $\begingroup$ That's awesome, thanks a lot! $\endgroup$
    – Aman Gupta
    Commented Feb 8, 2019 at 10:14
  • $\begingroup$ Try to redo the calculations. That may show where I made a clerical error (I hope not, but you never know) and it helps you actually understand the solution. $\endgroup$
    – Ingix
    Commented Feb 8, 2019 at 10:19
  • $\begingroup$ And how does the equation change when I need only integer values for x1 (and eventually for all) variables? $\endgroup$
    – Aman Gupta
    Commented Feb 8, 2019 at 10:33
  • $\begingroup$ It doesn't change, at least if $m,n,o,p$ are also integers. All the equations to calculuate $x_2,y_1$ and $y_2$ will give you integer results if you start with $x_1$ as integer. The inequality conditions also don't change, it may just happen that if the resulting max/min values come from the halve fractions, you must start with the next higher/lower integer. So if e.g. for a given parameter set we get for real $x_1$ the lowe bound $x_1 \ge 17.5$, that would need to be changed to $x_1 \ge 18$, and similar for the $x_1 \le ...$ case. $\endgroup$
    – Ingix
    Commented Feb 8, 2019 at 10:42
  • $\begingroup$ Okay, I get it. That's cool. $\endgroup$
    – Aman Gupta
    Commented Feb 8, 2019 at 10:43

Above equation shown below:

$\begin{align} x_1 + y_1 &= m\\ x_2 - y_1 &= n\\ x_1 - y_2& =\phi\\ x_2 + y_2 &= p\end{align}$

Solution is: $(x_1,x_2,y_1,y_2)=[(4),(p+\phi-4),(m-4),(p-m-n+4)]$

Where, $(m+n)=(p+\phi)$

For $(m,n,\phi,p)=(7,3,2,8)$ we get:



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