Prove that if $a|b_1b_2$, $a|c_1c_2$ and $a|b_1c_1+b_2c_2$, then $a|b_1c_1$ and $a|b_2c_2$

Question

Prove that if $a|b_1b_2$, $a|c_1c_2$ and $a|b_1c_1+b_2c_2$, then $a|b_1c_1$ and $a|b_2c_2$.

Solution:

$a|b_1b_2$: $\exists$ $c$ that $ac=b_1b_2$

$a|c_1c_2$: $\exists$ $d$ that $ad=c_1c_2$

$a|b_1c_1+b_2c_2$: $\exists$ $e$ that $ae=b_1c_1+b_2c_2$

$ae=b_1c_1+b_2c_2=...$

How to get the result I want?

Thank You.

Answer 1

Swap $c_1$ with $c_2$. Let $$f(x)=(b_1x+c_1)(b_2x+c_2).$$ Your condition means that $f(x)$ has coefficients divisible by $a$, that is $f(x)/a$ has integer coefficients. By Gauss's lemma, as $f(x)/a$ has linear factors with rational coefficients, then it has linear factors with integer coefficients. These must be $(b_ix_i+c_i)/a_i$ with $a_1a_2=a$. Then $b_1c_2/a=(b_1/a_1)(c_2/a_2)\in\Bbb Z$ etc.

Answer 2

Okay, prime factorization time.

Let $p$ be a prime factor of $a$ then $p|b_1b_2$ so it must either be a factor of $b_1$ or $b_2$. Same thing for $c_1$ and $c_2$; $p$ most be a prime factor of one of those. And $p$ is a prime factor of $b_1c_1+b_2c_2$. If $p|b_1c_1$ the $p$ must divide $b_2c_2$. And vice versa. And as $p$ must be a divisor of $b_1$ or $b_2$ it must divide one of $b_1c_1$ or $b_2c_2$. So it must divide both.

So we have $p|b_1b_2$ and $p|c_1c_2$ and $p|b_1c_1$ and $p|b_2c_2$.

Let $p^{m_a}|a$ so that $p^{m_a + 1}\nmid a$ (we'll call that the highest power of $p$ that divides $a$). $m_{b_1}, m_{b_2}, m_{c_1}, m_{c_2}$ be the highest powers of $p$ that divide $b_1, b_2,c_1,c_2$.

And let's let $b_1', b_2', etc....$ all be the values so that $b_1 = p^{m_{b_1}}*b_1'$, etc.

So $p\nmid b_1'$ etc.

Now $p^{m_a}|b_1b_2$ so $m_a \le m_{b_1} + m_{b_2}$ and likewise $p^{m_a} \le m_{c_1} + m_{c_2}$

Either $m_{b_1} + m_{c_1} \le m_{b_2} + m_{c_2}$ or $m_{b_1} + m_{c_1} \ge m_{b_2} + m_{c_2}$. As the labels are arbitrary, let's assume $m_{b_1} + m_{c_1} \le m_{b_2} + m_{c_2}$ and we'll just switch the indexes $1$ and $2$ if its the other way.

Bear with me. Let's also assume without loss of generality that $m_{b_1} \le m_{c_2}$. (We'll just switch the $b$ and $c$ labes otherwise.)

So:

$p^{m_a}|b_1c_1 + b_2c_2 = p^{m_{b_1}}b_1'p^{m_{c_1}}c_1'+p^{m_{b_2}}b_2'p^{m_{c_2}}c_2'=$

$p^{m_{b_1} + m_{c+1}}[b_1'c_1' + p^{(m_{b_2}+m_{c_2})- (m_{b_1} + m_{c-1})}b_2'c_2']$

Now $p\nmid b_1'c_1'$ so either $p\nmid b_1'c_1' + p^{(m_{b_2}+m_{c_2})- (m_{b_1} + m_{c+1})}b_2'c_2'$ or $m_{b_2}+m_{c_2} = m_{b_1} + m_{c-1})$.

Assume the former:

Then $m_a = m_{b_1} + m_{c-1}$ and $p^{m_a}|b_1c_1$ and $p^{m_a}|b_2c_2$.

Assume that latter: and assume $p|b_1'c_1' + b_2'c_2'$

Then $m_a > m_{b_1} + m_{c_1} = m_{b_2} + m_{c_2}$. But $a_m \le m_{b_1} + m_{b_2} \le m_{c_2} + m_{b_2}$ so that is a contradiction.

So $p^{m_a}|b_1c_1$ and $p^{m_a}|b_2c_2$.

Inductively do that for each prime factor.

Answer 3

Since $a\mid(b_1c_1+b_2c_2)$, we have $b_1c_1=am+r$ and $b_2c_2=an-r$ for some remainder $r$. But now $a^2\mid b_1b_2c_1c_2=a^2mn+ar(n-m)-r^2$ implies $a\mid r^2$, from which inductively we can conclude $a^{n-1}\mid r^n$ for all $n\ge1$. But that's not possible if $1\le r\lt a$. So we must have remainder $r=0$, i.e., $a$ divides $b_1c_1$ and $b_2c_2$.