Find all integers $x,y,z$ such that $6x+10y+15z=-1$.

Question

I did this-

$6x+5(2y+3z)=-1$

$6(x+1)+5a=5$ ($a=2y+3z$)

One solution to above equation is $x+1=5,a=-5$.

General solution would be $x+1=5+5k,$ $a=-5-6k$ where $k$ is an integer.

Now $2y+3z=-5-6k$. I don't how to go ahead from here. Any hint?

Source- Challenges and thrills of pre-college mathematics.

since $(2,3)=1$ so for any integer $k$ we can solve $2y+3z=-5-6k$ (and there are countless $(y,z)$ pairs for the solution..) — VicaYang, Sep 01 '19 at 07:49
So you are saying that solution for $y,z$ will depend on $k$ and one more integer(?) We need to find a general solution. — Jayant Jha, Sep 01 '19 at 07:57
Solve your equation when $k=1$, then when $k=2$, then when $k=3$, and so on. Try to spot or derive a pattern. — preferred_anon, Sep 01 '19 at 08:04
You want $2y+5$ is divisible by $3$ (or $3z+5$ is divisible by $2$). Choose one of these for the next step. — Mark Bennet, Sep 01 '19 at 08:14

Reza Rajaei · Accepted Answer · 2019-09-01T14:05:27.000

You can generate all solutions to this equation in the way explained below:

Clearly, $z$ is an odd integer. So assume $z=2k+1$, in which $k$ is an integer as well.

After re-writing the equation, $3x+5y+15k=-8$ therefore $3|8+15k+5y$. So, $3|2y+2$. It means that $y=3m+2$, in which $m$ is an integer too.

At the end of the computation, $3x=-8-15k-15m-10$, and $x=-5m-5k-6$. So,

$$(x,y,z)=(-5m-5k-6,3m+2,2k+1)$$

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