Diophantine equation $9x_1$+$12x_2$+$16x_3$=13

Question

Find all integral solutions of the linear Diophantine equation $9x_1$+$12x_2$+$16x_3$=13

The hint given said to solve the Diophantine equation $3y+16x_3$=$13$, so I found the integral solutions of that Diophantine equation to be $y=-1+16n$ and $x_3=1-3n$.

Then it said to use the general value of $y$ obtained to solve $9x_{1}+12x_{2}=3y$, but I don't see how to do that. Plugging in the obtained $y$ gives $9x_{1}+12x_{2}=3(-1+16n)$, and I can't seem to solve that.

Answer 1

One way to approach this problem is with the theory of vectors.

Let $(a,b,c)$ repeesent an ordered triple where $a=x_1,b=x_2,c=x_3$. Define $n(a,b,c)=(na,nb,nc)$.

Clearly one solution is $(1,-1,1)$. Other solutions must then satisfy the superposition relation

$(x_1,x_2,x_3)=(1,-1,1)+(x_1^*,x_2^*,x_3^*)$

where the second term satisfies the homogeneous relation

$9x_1^*+12x_2^*+16x_3^*=0$

The solutions to the homogeneous equation form a two-dimensional vector space, they are orthogonal to $(9,12,16)$. So it is sufficient to find two linearly independent solutions. One is obtained by looking at the case $x_3^*=0$:

$9x_1^*+12x_2^*=0, (x_1^*,x_2^*,x_3^*)=n_1(4,-3,0)$

A second linearly independent soution is obtained from the case $x_1^*=0$:

$12x_2^*+16x_3^*=0, (x_1^*,x_2^*,x_3^*)=n_2(0,4,-3)$

Now superimpose the two independent homogeneous solutions with the partucular solution $(1,-1,1)$ and get:

$x_1=1+4n_1$

$x_2=-1-3n_1+4n_2$

$x_3=1-3n_2$.

Answer 2

The hint given said to solve the Diophantine equation $3y+16x_3$=$13$, so I found the integral solutions of that Diophantine equation to be $y=-1+16n$ and $x_3=1-3n$.

You are correct.

Then it said to use the general value of $y$ obtained to solve $9x_{1}+12x_{2}=3y$, but I don't see how to do that.

Note that we can divide the both sides of $9x_1+12x_2=3y$ by $3$ to obtain $$3x_1+4x_2=y\tag1$$ Since $3\cdot (-1)+4\cdot 1=1$, multiplying the both sides of $3\cdot (-1)+4\cdot 1=1$ by $y$ gives $$3\cdot (-y)+4\cdot y=y\tag2$$ Now $(1)-(2)$ gives $$3(x_1+y)+4(x_2-y)=0$$ Since $3$ is coprime to $4$, we can write $$x_1+y=4m,\quad x_2-y=-3m,$$ i.e. $$x_1=4m-y=4m-16n+1,\quad x_2=-3m+y=-3m+16n-1$$

So, we get $$(x_1,x_2,x_3)=(4m-16n+1,-3m+16n-1,-3n+1)$$ where $m,n\in\mathbb Z$.