Finding integer solutions for $6x+15y+20z=1$ Problem: Find integers $x$, $y$, and $z$ that satisfy the equation $6x+15y+20z=1$.
I noticed that $\gcd(6,15)=3$, $\gcd(15,20)=5$, and that $\gcd(6,20)=2$. And of course $\gcd(6,15,2)=1$. 
Of course I know to set one of the variables (x, y, or z) to be zero. But I want to know how to get the more trivial answers. But where do I go from here to find the solutions?
 A: Note that there is an obvious solution $\ 6 + 15 - 20 = 1.\,$ More algorithmically, we can use the method of the Extended Euclidean Algorithm to compute $\rm\,gcd(6,15,20) = 1\,$ in a couple steps
$$\begin{array}{rrr}
[1]&\  20&  0&  0&  1\\
[2]&\  15&  0&  1&  0\\
[3]&\  6&  1&  0&  0\\
[1] -[2]\, =\, [4]& 5& \!\!0&  -1&  1\\
[3] -[4]\, =\,[5]& 1& 1&  1& \!\!\!\!-1
\end{array}\qquad\qquad\qquad\quad$$
where the row $\ n\,\ a\,\ b\,\ c\,\ d\ $ denotes that $\ n = 6a + 15 b + 20 c.\ $ Thus the final row yields
$$\quad 1 =  6 + 15 - 20$$
A: $$6x+15y+20z=1\iff5\,(x+3y+4z)+x=1\iff x=5a+1$$
$$6x+15y+20z=1\iff2\,(3x+7y+10z)+y=1\iff y=2b+1$$
$$6x+15y+20z=1\iff3\,(2x+5y+7z)-z=1\iff z=3c-1$$
$$6x+15y+20z=1\iff(30a+6)+(30b+15)+(60c-20)=1\iff a+b+2c=0$$

$\quad\to a\equiv b\mod 2\to$ Pick any two numbers a and b of the same parity (i.e., either both are even or both are odd), let $c=-\dfrac{a+b}2$ , then compute the values of x, y, and z according to the above $3$ formulas, and you're done ! :-)
A: Hint: Using Euclidean Algorithm you can find the gcd of the 3 numbers. Try using Euclidean Algorithm in the inverse direction some way.
gcd(a, b, c) = gcd(gcd(a, b), c)
$15=6 \times 2+3$ so $3= 1 \times 15-6 \times 2$ (+)
$20=6 \times 3+2$
$3=2 \times 1+1$
$1=3-2=3-20+6 \times 3=-20+7 \times 3=-20+7 \times 15-14 \times 6$ using (+)
So $z=-1$,$y=7$,$x=-14$ is a solution
Hope this helps.
A: Find a solution such as $(x,y,z)=(6,-1,-1).$
So, we have 
$$6x+15y+20z=1$$
$$6\cdot 6+15(-1)+20(-1)=1.$$
Hence, we have
$$6(x-6)+15(y+1)+20(z+1)=0.$$
This is much easier to treat.
A: Of course I know to set one of the variables (x, y, or z) to be zero. But I want to know how to get the more trivial answers.
No, you don't.  If you do that you will get $aw + bu = 1$ where $\gcd(a,b) \ne 1$ which is impossible.  In particular $6x + 15y = 1; 6x + 20z= 1; 15y + 20z = 1$ which are all impossible.
But if you set one of the variables, say $z$, to $20z \equiv 1 \mod \gcd(x,y)$ you will get $6x + 15y = k*\gcd(6,15)$ which will be solvable.
So $20z \equiv 1 \mod \gcd(6,15) \equiv 1 \mod 3$
e.g. $z = 2$ so $6x + 15y + 40 = 1$ and $2x + 5y = - 13$ which can be solved with $x = 1; y= -3; z=2$
So how to get all the other solutions?  (Did you really mean to say "more trivial"?)
Will if you fix one variable, say $x$, we know $6(x - 5*k) + 15(y + 2*k) + 20z = 6x + 15y + 20z$.  
We can generalize this to $6(x - 5*k - 10*j) + 15(y + 2*k - 4*l) + 20(z + 3k + 3*l) = 6x + 15y + 20z$.
(Or more general  $a(x - \frac{\text{lcm}(a,b)}{a} - \frac{\text{lcm}(a,c)}{a}) +b(y + \frac{\text{lcm}(a,b)}{b} - \frac{\text{lcm}(b,c)}{b})+c(z + \frac{\text{lcm}(a,c)}{c} + \frac{\text{lcm}(b,c)}{c})=ax + by+ cz$.)
So solultions are $x = 1 - 5k - 10j; y = -3 + 2k - 4l; z = 2 + 3j+3l$ will be all solutions for any integer $j,k,l$.
A: Put 
$$x=-5u-10v-9\\
y=\;\;2u\;\;\;\;\;\;\;\;+1\\
z=\;\;\;\;\;\;\;\;\;\;\;3v+2$$
This gives
$$6x+15y+20=6(-5u-10v-9)+15(2u+1)+20(3v+2)=1$$
Hence solutions to $(x,y,z)$ can be generated by substituting different integer values of $u,v$.
