Full question : Let a, b and c be non-zero integers. Prove that if ax+by=c has an integer solution, then it has infinitely many integer solutions (x, y) where x is even or it has infinitely many integer solutions (x, y) where y is even.
I'm confused on how to prove this, using LDET1, i know that the gcd(a,b)=d and d must divide c.
Now if I take the first case where x is even in (x,y), i get 2 more cases, when y is even and y is odd case 1: i can put x=2k, y=2m and prove that d is still divisible by c, even after the substitution. case 2: putting x=2k and y=(2l+1) i get something like d(2+b/d)=c, since b/d is an integer (d is gcd of a,b) 2+b/d is also an int and d/c in this case too.
Am i doing this right?
If $ax+by=c$ for some $x$ and $y$, then for any integer $n$:
$a(x+nb)+b(y-na)=ax+abn+by-abn=ax+by=c$.
So, for any solution $(x,y)$, you can get infinitely many other solutions $(x+nb,y-na)$.
Note that this proof works whether $x$ or $y$ are even or not.