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For every integer n, if $2 | n$ and $3 | n$ then $6 | n$

! Note: x | y means y is divisible by x.

!! Note: I know that there are way better ways to prove it. However, I am just curious whether the proof below, admittedly peculiar, is correct.

Since 2 | n and 3 | n, we can write $\frac{n}{2} = x $ and $\frac{n}{3} = y$ where $x,y \in \mathbb Z$. Therefore $$\tag1 \frac{n}{2} + \frac{n}{3} = x + y$$ $$\tag2 \frac{5n}{6} = x + y $$ $$\tag3 5\cdot\frac{n}{6} = x + y$$

Since $x, y \in \mathbb Z$, it follows that $x + y$ in integer and $5\cdot\frac{n}{6}$ is integer is as well. Need to prove that $\frac{n}{6} \in \mathbb Z$.

Suppose $\frac{n}{6} \notin \mathbb Z$. Since $5 \cdot \frac{n}{6}$ is an integer, $\frac{n}{6}$ can be rewritten as $\frac{n}{6} = a + 0.2$, where $a \in \mathbb Z$. But then it will imply that $n = 6a + 1.2$, meaning that $n \notin \mathbb Z$, hence a contradiction. Therefore, $\frac{n}{6} \in \mathbb Z$

Is it correct?

Robert Lewis
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Stokolos Ilya
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3 Answers3

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Simpler: $\ \dfrac{n}2\in\Bbb Z,\, \dfrac{n}3\in \Bbb Z\,\Rightarrow\, \dfrac{n}6 = \dfrac{n}2-\dfrac{n}3\in\Bbb Z.\ $ Turning to your argument:

Suppose $\frac{n}{6} \notin \mathbb Z$. Since $5 \cdot \frac{n}{6}$ is an integer, $\frac{n}{6}$ can be rewritten as $\frac{n}{6} = a + 0.2$, where $a \in \mathbb Z$.

This claim is unfounded.

Remark $ $ More generally $\,a,b\mid n\iff {\rm lcm}(a,b)\mid n\ $ and this can be proved as above.

Bill Dubuque
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Proposition:

If

$p, q \in \Bbb P \tag 1$

are distinct primes, and

$n \in \Bbb Z \tag 2$

with

$p \mid n, \; q \mid n, \tag 3$

then

$pq \mid n; \tag 4$

for (3) by definition means

$\exists a, b \in \Bbb Z, \; n = pa = qb, \tag 5$

which implies

$p \mid qb; \tag 6$

now

$p \ne q \Longrightarrow \gcd(p, q) = 1, \tag 7$

and this by Bezout's identity yields

$\exists x, y \in \Bbb Z, \; xp + yq = \gcd(p, q) = 1, \tag 8$

whence

$xpb + yqb = b; \tag 9$

using (5) this may be written

$p(xb + ya) = xpb + ypa = b; \tag{10}$

this shows that

$p \mid b \Longrightarrow \exists c \in \Bbb Z, \; b = pc \Longrightarrow n = qb = pqc \Longrightarrow pq \mid n. \tag{11}$

QED.

Applying this proposition to the case at hand we take

$p = 2, \; q = 3, \tag{12}$

and thus conclude immediately that

$6 \mid n. \tag{13}$

$OE\Delta$.

Robert Lewis
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Your question is whether your proof is correct. Some comments have indicated that it is flawed, and some answers have shown different approaches, which you indicated you were not interested in. So here is how I would massage your proof to retain as best I can its line of reasoning.

When you get to the step $5\cdot \frac{n}{6}= x+y$, you do not deal with the consideration that if $\frac{n}{6} \in \mathbb Z$, it must be the case that $\frac{x+y}{5} \in \mathbb Z$. So better to use difference of terms rather than the sum as in Bill Dubuque's answer, and get directly to the question , is $\frac{n}{6} \in \mathbb Z$?

Then what you can say is that $\frac{n}{6}=a+\frac{(n\mod 6)}{6}$ where $a\in \mathbb Z$. In order for $\frac{n}{6} \in \mathbb Z$, it will have to be the case that $\frac{(n\mod 6)}{6} \in \mathbb Z$, and that only happens if $\ (n\mod 6)=0$, which is the very definition of $6\mid n$.

Keith Backman
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