Prove or disprove the following statements:
(a) If $d \mid a$ and $d \mid b$, then the Diophantine equation $ax + by= c$ has solutions.
(b) If $d \mid a$, $d \mid b$ and $d \mid c$, then the Diophantine equation $ax + by= c$ has solutions.
(c) The Diophantine equation $ax + (a + 1) y= 1$ has solutions.
(a) The equation $ax + by = c$ has integer solutions if $\gcd(a,b) \mid c$.
However, we have no reason to believe $\gcd(a,b)$ does indeed divide $c$.
We're given $d \mid a$ and $d \mid b$ but these have no constraint on $c$.
And so the statement is not always true.
(b) The equation $ax + by = c$ has integer solutions if $\gcd(a,b) \mid c$.
We're given $d \mid a$, $d \mid b$ and $d \mid c$, which means for some integers $p,q,r$,
$$ a = pd $$
$$ b = qd $$
$$ c = rd $$
This gives us
$$ \gcd(a,b) = \gcd(pd, qd) = d \gcd(p,q) $$
The equation has integer solutions if $\gcd(a,b) \mid c$, which is restated as
$$ d \gcd(p,q) \mid rd $$
That is, $\gcd(a,b)$ devides $c$ is there exists some integer $s$ such that
$$ rd = sd \gcd(p,q)$$
Simplified
$$ r = s \gcd(p,q)$$
There aren't enough constraints for us to determine this.
So in general the statement is not true.
(c) The Diophantine equation $ax + (a + 1) y= 1$ has integer solutions if $\gcd(a, a+1) \mid 1$.
This is only possible if $\gcd(a, a + 1) = 1$.
We know from exercise (1.1).18 that two consecutive integers are coprime.
So the equation has integer solutions.