Prime Numbers. Show that if $a \mid 42n + 37$ and $a \mid 7n +4$, for some integer $n$, then $a = 1$ or $a = 13$ Show that if $a \mid 42n + 37$ and $a \mid 7n +4$, for some integer $n$, then $a = 1$ or $a = 13$
I know most of the rules of divisibility and that any integer number can be expressed as the product of primes.
Given $a = \prod_{i=1}^\infty p_i^{\alpha_i}$ and $b = \prod_{i=1}^\infty p_i^{\beta_i}$
$a \mid b$ if and only of $\alpha_i \le \beta_i$ for every $i = 1, 2, 3, \ldots$
Even though i have this information, I cannot prove the statement. Help me please.

If $a$ divides $42n+37$ and $a$ divides $7n+4,$
then $a$ divides $42n+37-6(7n+4)=13$.  
Can you take it from here?
