Let $aThe prove I tried is the following. I really wish someone can check if I made some logical mistake, especially the last part I found myself diffident proving $a$ can only be $1$ or $5$.
Because $b\neq c$ and $b<c$, otherwise switch the value of $b,c$. Becuase $a\mid (3b+2c)$ and $a \mid (3c+2b), \exists q_1,q_2 \in N$ that $$aq_1=(3b+2c), aq_2=(3c+2b), a(q_2-q_1)=c-b.$$
Therefore we have $a\mid (c-b)$ and $a \mid 3(c-b)+(3b+2c)$ which is $a \mid 5c$. And we have $a \mid 2(c-b)+(3c+2b)$, $a\mid 5c$. Hence $a \mid 5b$ and $a \mid 5c$. Because $b,c$ are distinct odd primes. This is possible only when $a\mid b$ and $a \mid c$ or $a \mid 5$. The only number that divides two primes is $1$. And the two numbers that divides $5$ are $1$ and $5$. Hence the two possible values for $a$ are $1$ and $5$.
 A: You proof is fine, but you are over elaborating at certain points. Do you need $q_1,q_2$, for example? You do not seem to have used them.
Indeed, the simplest proof is that if $a|3b+2c$ and $a | 3c+2b$, then $a | 3(3c+2b) - 2(3b+2c) = 5c$, and $a | 3(3b+2c) - 2(3c + 2b) = 5b$.
Therefore, $a | 5c$ and $a | 5b$, which means that $a$ divides the greatest common divisor of $5c$ and $5b$. But, $(5c,5b) = 5(c,b) = 5 \times 1 = 5$ (as $c,b$ are prime), so $a | 5$.
You have done the same thing, but with a little more elaboration.
Also, it is useful to show that both cases are attained : for example, $b = 2,c=7$ gives $3b+2c = 20$, and $3c+2b = 25$.
However, note that if $5 | 3c+2b$ and $5 | 3b+2c$, then $5 | b-c$. Given that $b,c$ are primes, this forces one of $b,c$ to be even. Thus, $b = 2$ and $c = 7$ is forced.
Conclusion : if $b \neq 2, c \neq 7$, then in fact $a = 1$ is forced given $b < c$.

Try this more general question : given $ b < c$ natural numbers not necessarily prime, and integers $d,e$ , if $a | db+ce$ and $a | eb+cd$, what can you say about $a$?
