If $a\mid b$ and $a\mid c$ prove $a\mid a^2 + 3b - 2^{b} c$

Question

So I have this problem for homework so I'm not looking for an answer, just some help moving forward.

So far, I have recognized that

$$\begin{align*} a\mid b &\Rightarrow b = ak,\\a\mid c &\Rightarrow c = aj,\\a\mid a &\Rightarrow a = ah,\end{align*}$$ where $k, j, h\in\mathbb{Z}$.

From there I have $a + b + c = ak + aj + ah = a(k + j + h) = ai$ where $i\in\mathbb{Z}$.

So I have been able to prove that $a\mid a + b + c$ but I am lost as to how to move onward to prove the desired statement.

Any and all help or suggestions are very much appreciated!

Answer 1

Hint.

What you've done with $a+b+c$, you can do with $Ka+Lb+Mc$, for any integers $K$, $L$, $M$, right?

Now choose $K=a$, $L=\ldots$

Answer 2

Strong Hint:

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$(*)$ Since $a\mid b$ then $a\mid 3b$. And, since $a\mid c$ then $a\mid 2^bc$.

It is trivial that $a\mid a^2$ so we cancel that out. $$a\mid\underbrace{a^2 + 3b + 2^bc}_{1} \longrightarrow a\mid \underbrace{3b + 2^bc}_{2}$$ Let $(2) = ak$, then if we add $a^2$ to it to equal $(1)$, we notice that $(1) = a(k + a)$ and is divisible by $a$. This is why we cancel out the term $a^2$, and so we know that the new expression above, namely $(2)$, is correct.

But also, looking at the statement $(*)$ highlighted in bold, we can draw an obvious conclusion :)

Answer 3

1) $a|a$. Always.

Pf: $a = a*1$.

2) If $a|b$ then $a|kb$ for all integers $k$

Pf: If $a|b$ then there is an integer $m$ so that $b = am$ so $kb = a(mk)$. $j*k$ is an integer so $a|kb$

3) If $a|b$ and $a|c$ then $a|b+c$

Pf: $a|b$ means there is an integer $k$ so that $b = ak$. $a|c$ means there is an integer $m$ so that $c = am$ so $b+c = ak + am = a(k+m)$. $(k+m)$ is an integer so $a|b+c$.

So... put that all together.

a) $a|a$ so $a|ka$ for all $k$ so $a|a*a=a^2$.

b) $a|b$ so $a|kb$ for all $k$ so $a|3b$.

c) $a|c$ so $a|kc$ for all $k$ so $a|-2^b*c$.

d) $a|a^2$ and $a|3b$ so $a|a^2 + 3b$.

d') $a|a^2 + 3b$ and $a|-2^b*c$ so $a|a^2 + 3b - 2^b*c $