Some doubt on Linear Diophantine equation We know $ax+by=c$ is solvable iff $(a,b)|c$ where $a,b,c,x,y$ are integers.
If $x=2$, $a=\dfrac{k(k+5)}{2}$, $y=k$, $b=k+3$ and $c=2k$, where $k$ is any integer, then
$$2 \frac{k(k+5)}{2} - k(k+3) = 2k.$$
So,  $\left(\dfrac{k(k+5)}{2}, (k+3)\right) | 2$ for all integral value of $k$.
But $k+3$ cannot be even unless $k$ is odd.
Where I am going wrong?

$2k(k+5)/2 - k(k+3)=2k$ with $a=k(k+5)/2$, $x=k$, $y=k$, would mean $b=k+3$, not $k(k+3)$.
The conclusion is that $(k(k+5)/2,k+3) | 2k$, not $2$.  For example, with $k=3$, $(24,6)=6$. 
