# generating solutions to diophantine equation

If I have a solution to say $$4x_1+3x_2+3x_3+3x_4...+3x_n=4y$$ where $$x_i ,y$$ are non negative integers can I generate other solutions from my initial solution? And if I can would all solutions follow from this initial one?

• what I am really trying to get at is if my initial solution contains all even integers then would solutions generated contain at least one even integer because the coefficient in front of $x_1$ is even – argamon May 15 at 0:40
• Is it a linear equation you are interested in? If yes, you sould find more solution by finding integer vectors in the "kernel" of this system. Here, it is the space orthogonal to $(4, 3, 3, \dotsc, 3)$ or more precisely the integer lattice living in this orthogonal. – KeiOh May 15 at 2:15
• For your comment, I am not sure it works: take $n=5$, and consider $u=(-3,1,1,1,1)$. Then $4u_1 + 3u_2 +\dotsb + 3u_5 = 0$. Now look at the equation $4x_1 + 3x_2 + \dotsb + 3x_5 = 8$. Then $x = (2, 0, \dotsc, 0)$ is a solution, but then so is $x+u$, isn't it? But $x+u$ as all odd coordinates. – KeiOh May 15 at 2:19

I'm not sure what exactly are you asking for, but I observed the following properties.

I'm assuming you want solutions for $$(x_1,\dots,x_n)$$ given $$y$$. Let $$z=y-x_1$$, then $$3(x_2+x_3+x_4\dots+x_n)=4z$$

Hence, $$z$$ must be a multiple of $$3$$, so we write $$3k=y-x_1\implies x_1=y-3k$$ for $$0\le k\le\left\lfloor\frac{y}{3}\right\rfloor$$.

If $$x_1=y-3k$$ for some $$0\le k\le\left\lfloor\frac{y}{3}\right\rfloor$$, then $$x_2,\dots,x_n$$ are a solution $$\iff\sum_{i=2}^n x_i = 4k$$.

Now to get all solutions, simply iterate all $$k$$, and for each, all $$x_2,\dots,x_n$$ s.t. $$\sum_{i=2}^n x_i = 4k$$.

That is, you can generate a solution form some other in two ways ("one two-case operation"):

1. (We redistribute the sum condition:) You can decrease (increase) some component $$x_i,i\ge 2$$ by $$1$$ as long as you increase (decrease) some other component $$x_{j},j\ge 2$$.

2. (We change the case of $$k$$:) You can decrease (increase) $$x_1$$ by $$3$$, as long as you increase (decrease) up to four other components $$x_i,x_j,x_t,x_l$$ for $$2\le i \le j \le t \le l$$ by $$1$$.

The set of all solutions is closed relative to this "two-case operation", so by repeating these two cases on any solution and all subsequently generated solutions, you can eventually reach all solutions.