3
$\begingroup$

Determine all triples $(x,y,z)$ of integers satisfying the equation $3x+4y+5z=6$

I am not familiar with Diophantine equations with more variables. How do I solve this? Please anyone suggest some easier way to solve these linear Diophantine equations.
I think this can help here -
If $(x_0, y_0)$ are solutions to $ax + by = C$ with $gcd(a,b) = 1$, then $(x_0 + bt, y_0 - at)$ for all integer $t$ are also solutions.

1 Answers 1

5

Note that we have that $$3x+4y=6-5z$$ So from what you know, we have that $(x,y)$ is characterized by $(-2+z-4t,3-2z+3t)$ for some integer $t$. So $(x,y,z)$ are for $t,z \in \mathbb{Z}$, $$(-2+z-4t,3-2z+3t,z)$$

  • 1
    Explain a little bit more... how to get the initial solution $(x,y)$ ? – 2017-02-03
  • 2
    @RezwanArefin Well, to find the initial solution we separate $x$ and $y$ into their $z$ parts, so $x=x_{0}+zx_{1}$, $y=y_{0}+zy_{1}$ that $3x_{0}+4y_{0}=6$ and $3x_{1}+4y_{1}=-5$. Using this, we can get the initial solution. – 2017-02-03
  • 0
    With another approach I got $(1 + 3s + 4t, 1 -s-3t, 1-s), s,t \in \mathbb{Z}$. Is it right also? The approach uses the fact that $3x + 4y \equiv 1\; \text{mod}\; 5$ or $3x + 4y = 1+5s$ – 2017-02-03
  • 0
    @Rezwan Well, that does follow from my characterization, but Yes, that works as well. – 2017-02-03