Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I’m having trouble with the proving the following:

If $a \mid b-2c$ and and $a \mid 2b+3c$ then $a \mid b$ and $a \mid c$.

Heres my partial solution:

By definition of divisibility, $b-2c=ak$ (for some $k\in\mathbb{Z}$) and $2b+3c=am$ (for some $m\in\mathbb{Z}$)

Then $b=ak+2c$: substituting into the other equation yields $2(ak+2c)+3c=am$. This means $2ak+7c=am$

I’m not sure where to go from here. Any help would be great :)

share|improve this question
    
Provided $(a, 7)=1$. Then it is correct! –  PAD Jul 13 '12 at 21:42
add comment

4 Answers

up vote 3 down vote accepted

Rearrange the equation $2ak+7c=am$ to get $7c=a(m-2k)$, showing that $a\mid 7c$. Note that while this would certainly be true if $a\mid c$, it would also be true if $a=7$, regardless of what $c$ might be. This suggests that we should go back to the original question and see what happens when $a=7$. In that case we have $7\mid b-2c$ and $7\mid 2b+3c$. Can this be true without having $7\mid b$ and $7\mid c$?

Equivalently, is there a solution other than $b\equiv c\equiv 0\pmod7$ to the system

$$\left\{\begin{align*}&b-2c\equiv 0\pmod7\\&2b+3c\equiv 0\pmod7\;?\end{align*}\right.$$

Mod $7$ the second congruence is actually just twice the first, since $-4\equiv 3\pmod 7$. Thus, the congruences are not independent, and any solution to the first must also be a solution to the second. It’s not hard to find a non-zero solution to the first congruence.

share|improve this answer
add comment

Hint $\rm\ mod\ a\!:\, b\equiv 2c,\ 3c\equiv -2b \equiv -4c\,\Rightarrow\, 7c\equiv 0.\,$ So $\rm\, \rm\color{#C00}{if\ \ 7\nmid a\ \ then}\,$ $\rm\,c\equiv0\,\Rightarrow\,b\equiv 2c\equiv 0.$

$\rm\color{#C00}{It\ fails\ if\,\ 7\:|\:a.}$ If $\rm\:a = 7c,\,$ let $\rm\:b = 2c.\:$ So $\rm\:a\:|\:b\!-\!2c = 0,\ a=7c\:|\:2b\!+\!3c = 7c,\,$ but $\rm\:a=7c\nmid c.$

Remark $\ $ Geometrically the result says that, modulo $\rm\:a,\:$ the lines $\rm\: y = 2x,\ 2y = -3x\:$ intersect only at the origin if $\rm\:7\nmid a,\:$ but otherwise they intersect elsewhere too, e.g. mod $7$ the lines coincide since $\rm\:2y = - 3x = 4x,\,$ so $\rm\:y = 2x\:$ by scaling by $4 \equiv 1/2$. Thus, being identical, their intersection is the entire line $\rm\:y = 2x,\:$ i.e. the $7$ points $\rm\:(x,2x)\ mod\ 7.$

If you know linear algebra, note that the system $\rm\:2x-y = 0 = 3x+2y\:$ has determinant $7.\,$ So if $7$ is coprime to $\rm\,a,\,$ then $7$ is invertible mod $\rm\,a,\,$ so we can use Cramer's rule to infer $\rm\:(x,y) = (0,0).$

share|improve this answer
add comment

When you get to $b−2c=ak $ and $2b+3c=am$, the next thing to do is to see if you can eliminate $b$ between these equations. That should allow you to show that $c$ is an expression with a factor of $a$. I will explain more if you need more of a hint.

Edit: Ah - I was assuming that the question itself was correct!

share|improve this answer
add comment

This is not always true. For example, let $a=7$, $b=9$, $c=1$.

One can show that the only counterexamples involve $a$ being a multiple of $7$.

Remark: One view of things is that the reason for the failure is the fact that the determinant of the matrix $$\begin{pmatrix} 1 &-2\\2 &3\end{pmatrix}$$ is divisible by $7$. In a problem of this type, we can conclude the divisibility only when the determinant of the matrix is equal to $\pm 1$.

For example, if we replace $b-2c$ by $b+2c$, then the desired divisibility result will hold. So a slightly different question may have been intended.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.