# Let $a<b<c$, where $a$ is a positive integer and $b$ and $c$ are odd primes. Prove that if $a \mid (3b+2c)$ and $a \mid (2b+3c)$, then $a=1$ or $5$.

The 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 $$b, otherwise switch the value of $$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$$.

• Your punctuation is a bit off but, other than that, this looks okay. Well done. – Shaun Dec 7 '18 at 6:03
• @Tianlalu, this is elementary number theory. The number-theory tag is mostly redundant once the elementary-number-theory tag is present; besides, the former is more concerned, for instance, with techniques from analysis. – Shaun Dec 7 '18 at 6:07
• @Shaun , got it. – Tianlalu Dec 7 '18 at 6:08

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 $$a$$?