$ax+by=d$ has integer solutions then $ax+by+cz=d$ has integer solutions

Question

the questions asks us to either prove or disprove the statement let $a,b,c \in \Bbb{Z}$ if $ax+by=d$ has integer solutions then $ax+by+cz=d$ has integer solutions. Is it enough to provide a counter example to disprove it?

Answer 1

If $ax+by =d$ has a solution $(x_0,y_0)$ then $ax +by +cz =d$ has a solution $(x_0,y_0,0)$

Clearly the other way conclusion doesn't hold. If we take $3x+6y+z=10$ which has solution $(1,1,1)$ the equation $3x+6y=10$ doesn't have a solution, since $3\nmid 10$.

Answer 2

$ax+by=d$ has solutions iff $\gcd(a,b)\mid d$

$ax+by+cz=d$ has solutions iff $\gcd(a,b,c)\mid d$

But since $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$ then $\gcd(a,b,c)\mid\gcd(a,b)\mid d$

So if the first equation has solutions, the second one has too.