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If $6|n$ then $n=6k=3k+2k+k$. And $3k|n$, $2k|n$ and $k|n$.

Now let $p,q$ and $r$ be three distinct divisors of $n$ so that :

$$n=p+q+r$$

Because $p|n $, $ q|n$ and $r|n$ I figured that $p|q+r $, $ q|p+r$ and $r|p+q$.

I tried to prove that n is even or a multiple of 3 but without much luck.

How can I prove the statement?

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  • $\begingroup$ I think the converse is not true. Take any three odd primes and sum them up. (say $3, 5, 7$). Their sum is an odd number and never divisible by $6$. $\endgroup$ – user1952500 Apr 22 at 21:31
  • $\begingroup$ Yes, but their sum ($15$ in your example) being $n$ is not divisible by all three of the alleged divisors ($7$ in your example does not divide $15$). $\endgroup$ – Mark Fischler Apr 22 at 21:52
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Your first line demonstrates that $$6|n \implies \exists (a,b,c) : (n = a + b + c) \wedge (a|n \wedge b|n \wedge c|n)$$ which is the "only if" part -- six divides $n$ only if $n$ can be written as the sum of three divisors. So we must now take care of the "if" part.

So assume (hypothesis of the "if" clause") $\exists (a,b,c) : (n = a + b + c) \wedge (n = pa \wedge n = qb \wedge n = rc)$. Then $$ \frac{n}{a} = 1 + \frac{b}{a}+ \frac{c}{a} = 1+ \frac{n}{qa}+ \frac{n}{ra} \\ \frac1{a}=\frac{1}{n}+\frac{1}{qa}+\frac1{ra}=\frac{1}a \left( \frac{1}{p}+\frac{1}{q}+\frac1{r} \right) \\ \frac{1}{p}+\frac{1}{q}+\frac1{r} = 1 $$ with $p,q,r$ positive integers.

Then at least one of $(p,q,r)$ must be less than or equal to $3$ because $\frac14+\frac14+\frac14 < 1$. One might consider $p=q=r=3$, but then $n$ does not have three distinct divisors. Sb one of them needs to be $2$; without loss of generality we can take $p=2$. Then $$\frac{1}{2}+\frac{1}{q}+\frac1{r} = 1 \implies \frac{1}{q}+\frac1{r} =\frac12 $$

So at least one of $(q,r)$ must be less than or equal to $4$ because $\frac15+\frac15 < \frac12$. $q = \frac14$ does not work because that would give $r = \frac14 = q$ and again they are not distinct. Thus $q=3$, and that implies that $r=6$. So $n = 2a = 3b = 6c$. But $n = 6c \implies 6|n$ which is what we were trying to prove.

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Without, loss of generality, presume that $p > q > r$. Observe that $$p|(q+r) \implies (q+r) = k_1 p \implies 1 \le k_1 = \frac{q+r}{p} < 2 \implies k_1 =1.$$ As such, $p = q+r$, and $n = 2(q+r)$. Since $q|n=2(q+r)$, we have $q|2r$. That is, $$2r = k_2 q \implies 1 \le k_2=\frac{2r}{q}<2 \implies k_2 = 1.$$ Therefore, $$q = 2r \implies p = q+r =3r \implies n = 6r.$$

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