when exactly does $711 \equiv \alpha \times 2015 \mod 1302$ have a solution? I know this is quite a stupid question, but I just can't figure out whether or not $$711 \equiv \alpha \times 2015 \mod 1302$$ has a solution for $\alpha \in \mathbb{Z}$.
In my eyes, 2015 and 1302 should be coprime in order for this equation to have a solution, but as the question has been asked to me, there should be an answer. This already confused me too much.
Can anybody help me please?
 A: $711 - \alpha 2015=1302x$ where $x \in Z$. So, Solve diphantine equation : $2015y + 1302x = 711$. Do you need more explanation?
$x \equiv y$ (mod $n$) means $x-y$ is divisible by $n$. So, $x-y=na$, $a \in Z$. I used this simple definition. And you can solve linear equations in integers. Such equations are called diophatine equations.
You can refer Burton's elementary number theory, for solution of linear Diophantine equations.
A: $2015 x \equiv 711\pmod{1302}\,$ means $\, 2015 x + 1302 y = 711\,$ for some integer $y.\,$ This implies that $\,d = \gcd(2015,1302)\,$ divides $711.\,$ Instead of using the Euclidean algorithm to compute $d$, notice $1302$ has obvious factors $\,2\cdot3 = 6$ coprime to $2015$. The cofactor $1302/6 = 217$ has obvious factor $7$ coprime to $2015,\,$ but its cofactor $217/7 = 31\mid 2015,\,$ but $31\nmid 711,\,$ contradiction.
Remark $\ $ This necessary condition for solvability is also sufficient, since then we can cancel the gcd $d$  to get an equation of the form $\, a x + b y = c\,$ for $\,a,b\,$ coprime. By Bezout's identity for the gcd there are $\,\bar x,\bar y\,$ with $\,a\bar x + b\bar y = 1 = \gcd(a,b).\,$ Scaling that by $cd\,$ yields the sought solution.
A: Observe that both 
$$\begin{cases}31\mid1302\\31\mid2015\end{cases}\implies 711=31a\cdot65+31k\cdot42$$
But $\;31\nmid 711\;$ , so...
A: Hint: Try to solve $2015 x+1302 y=711$ in integers.
