generating solutions to diophantine equation

Question

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?

Answer 1

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

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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.