How to prove whether these two equations are satisfied at the same time?

Question

I want to prove whether both $2p-m$ and $p+m$ are the multiples of $3$, where $p,m \in \mathbb{Z}$.

My solution:

Assume $2p-m = 3k$ and $2q-n = 3l$, where $k,l \in \mathbb{Z}$. So I could get $m = 2l-k$ and $p = k+l$. But I do not know how to prove whether above conditions must be satisfied at the same time. Could anyone help me out?

More specifically, if $p$ and $m$ are integers, $p \in [-N,N]$ and $m \in [-M,M]$. Is the answer the same? Thanks in advance!

Answer 1

What is your question exactly?

1) $2p - m$ and $p +m$ are either both divisible by $3$ or neither are.

Divide $2p - m$ by three and you get a quotient, call it $q_1$ and you get a remainder, call it $r_1$, so that $2p -m = 3q_1 + r_1$

Divide $p+m$ by three and you get a quotient, call it $q_2$ and you get a remainder, call it $r_2$, so that $p+m = 3q_2 + r_2$

So $(2p - m) + (p+m) = 3p$ is divisible by three. But $(2p-m) + (p+m) = (3q_1 + r_1) + (3q_2 + r_2) = 3(q_1 + q_2) + (r_1 + r_2)$. So either $r_1 + r_2=0$ or $r_1 + r_2 = 3$.

If $r_1= 0$ then $r_2=0$ and both are divisible by $3$.

If $r_1= 1$ or $r_1 = 2$ then $r_2 =2$ or $r_1 = 1$ and neither are divisible by $3$.

Was that your question?

2) If $2p - m$ and $p +m$ are both divisible by $3$ then the remainder of $p$ when divided by $3$ is the "opposite" of the remainder when $m$ is divided by $3$, in other words, if the remainder of $p$ is $k$ then the remainder of $m$ is $j= 3-k$ if $k \ne 0$ or it is $j = k = 0$ if $k = 0$.

If $m = 3q_3 + r_3$ and $m= 3q_4 + r_4$ then $2p -m = 2(3q_4 + r_4) - (3q_3 + r_3)= 3(2q_4 - q_3) + (2r_4 -r_3)$ so $2r_4 - r_3 = 0$ or $2r_4 - r_3 = 3$

And $p + m = 3q_4 + r_4 + 3q_3 + r_3 =3(q_4 + q_3) + (r_4 + r_3)$ so $r_4 + r_3 = 0$ or $r_4 + r_3 = 3$.

If $r_4 = 0$ then $r_3 = 0$.

If $r_4 =1$ then $r_3=2$.

If $r_4 = 2$ then $r_3 =1$.

Was that your question?

Answer 2

I do not get exactly your question but it could only be these two cases:

a. Are you asking how to set up satisfying equation that generalize those two equations (assuring that both were satisfied simultaneously)?

Well if that is the case then just do the modular arithmetic:

$$(2p-m)(p+m) (mod 9) = 0$$

b. Are you asking how to set up a solution so that both relation holds?

Well you also make use of modular arithmetic. First let us assume that the first relation holds. That is: $$(2p-m) (mod3) = 0$$

Implying that, $$2p (mod3) = m$$

Now we multiply the second relation $p+m$ by 2 to get: $$2(p+m)$$ and this new relation should also be divisible by 3: $$(2p + 2m) (mod3)= (m+2m)(mod3) = 3m (mod3) = 0$$

Hence, whenever $2p-m$ is divisible by $3$, so thus $p+m$!