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$a,b,c$ are positive, distinct, co-prime integers. If all $a,b,c$ divides $a+b+c$, then $a+b+c=6$. Co-prime does not mean pairwise co-prime in this text, but that $gcd(a,b,c)=1$.

Found it out while testing my tool BigZ and leave it on ME for someone who want to prove it.

Is it okay to use MSE as a "proof calculator"?

Order the numbers as $a<b<c$.

Now $c,|,a+b+cimplies c,|,a+bimplies a+b=nc$ for some $nin N$. But clearly $a+b<2c$ so we must have $n=1$ and $a+b=c$

Note: if $gcd(a,b)=d$ then $d$ also divides $c$, so relative primality tells us that $d=1$

The assumption $a,|,a+b+c$ now tells us $a,|,2a+2bimplies a,|,2b$. Since $gcd(a,b)=1$, it follows that $a,|,2$ so $ain {1,2}$.

Case: $a=1$. Then we have $b+1=c$ and $b,|,c+1$ Thus $b$ divides both $c-1$ and $c+1$ but the gcd of those terms is $2$. Hence $b=2$, $c=3$ and $a+b+c=6$ as desired.

Case: $a=2$. Then $b+2=c$ and $b,|,c+2$. Then $b$ divides both $c-2$ and $c+2$ so $b$ divides $4$. But that is impossible (since $b>a$ this would imply $b=4$ and $c=6$, but $(2,4,6)$ are not mutually coprime).

Suppose $a> b> c$ (we can do that because of symmetry, equality is excluded because they are coprime). Then $$amid b+c implies kcdot a = b+c <2a implies k<2$$
So $k=1$ and now we have $b+c=a$. So $$ bmid a+b+c =2aimplies bmid 2$$ and $$ cmid a+b+c =2aimplies cmid 2$$
So $c=1$ and $b=2$ and thus $a=3$.

It is also not too difficult to solve for $(a,b,c)inmathbb{Z}_{neq 0}^3$ such that $a$, $b$, and $c$ divides $a+b+c$. Up to permutation, all solutions $(a,b,c)$ are given in the five infinite families below, i.e., the equations (1)-(5).

The trivial case is when $c=-a-b$, that is, we have an infinite family of solutions $$(a,b,c)=(a,b,-a-b),,tag{1}$$ where $a,binmathbb{Z}_{neq 0}$ are such that $a+bneq 0$. From now on, we assume that $a+b+cneq 0$.

Without loss of generality, we suppose that $|a|leq |b|leq |c|$. Hence, it follows immediately that $0<|a+b+c|leq |a|+|b|+|c|leq 3|c|$. As $c$ divides $a+b+c$, we can then conclude that
$a+b+cin{pm c,pm 2c,pm3c}$.

If $a+b+c=+c$, then $a+b=0$. Then, we have an infinite family of solutions $$(a,b,c)=(a,-a,na),,tag{2}$$ where $a,ninmathbb{Z}_{neq 0}$.

If $a+b+c=-c$, then $a+b=-2c$. Since $a$ and $b$ must divide $c$, we conclude that $a=b=-c$. Consequently, we have solutions $(a,b,c)=(a,a,-a)$ for $ainmathbb{Z}_{neq 0}$. Nonetheless, these solutions are obtained by permuting the solutions in (2) with $n=1$.

If $a+b+c=+2c$, then $a+b=c$. Therefore, $a$ and $b=c-a$ are divisors of $a+b+c=2c$. Let $2c=ka$ for some integer $kneq 0$. Then, $b=c-a=c-dfrac{2c}{k}$. Since $dfrac{2c}{b}$ is an integer, we must have $dfrac{2c}{c-frac{2c}{k}}=dfrac{2k}{k-2}=2+dfrac{4}{k-2}$.
That is, $k-2mid 4$, so $kin{-2,0,1,3,4,6}$. Note also that $|a|leq |c|$ implies that $|k|geq 2$. Therefore, $kin{-2,3,4,6}$.

• The case $k=-2$ gives $a=-c$ and $b=2c$, yielding solutions $(a,b,c)=(a,-2a,-a)$ for $ainmathbb{Z}_{neq 0}$. However, these solutions violate the assumption that $|a|leq |b|leq |c|$.




• The case $k=3$ gives $c=3t$ for some $tinmathbb{Z}_{neq 0}$, along with $a=2t$ and $b=t$, yielding solutions $(a,b,c)=(2t,t,3t)$ with $tinmathbb{Z}_{neq 0}$. However, these solution violate the assumption that $|a|leq |b|leq |c|$.




• The case $k=4$ gives $c=2a$ and $b=a$, yielding solutions $$(a,b,c)=(a,a,2a)
  tag{3}$$ with $ainmathbb{Z}_{neq 0}$.




• The case $k=6$ gives $c=3a$ and $b=2a$, yielding solutions $$(a,b,c)=(a,2a,3a)tag{4}$$ with $ainmathbb{Z}_{neq 0}$.

If $a+b+c=-2c$, then $a+b=-3c$, but $3|c|=|a+b|leq |a|+|b|leq 2|c|$, leading to a contradiction. Therefore, there are no solutions in this case.

If $a+b+c=+3c$, then $a+b=2c$. Since $|a|leq |b|leq |c|$, this situation can occurs iff $a=b=c$, so we have solutions $$(a,b,c)=(a,a,a)tag{5}$$ for $ainmathbb{Z}_{neq 0}$.

If $a+b+c=-3c$, then $a+b=-4c$, but $4|c|=|a+b|leq |a|+|b|leq 2|c|$, leading to a contradiction. Therefore, there are no solutions in this case.

In particular, all solutions $(a,b,c)inmathbb{Z}_{>0}^3$ with $aleq bleq c$ are of the form $(a,b,c)=(a,a,2a)$, $(a,b,c)=(a,2a,3a)$, and $(a,b,c)=(a,a,a)$ with $ainmathbb{Z}_{>0}$. Among these positive-integer solutions, the triples $(a,b,c)$ with $gcd(a,b,c)=1$ are then $(a,b,c)=(1,1,2)$, $(a,b,c)=(1,2,3)$, and $(a,b,c)=(1,1,1)$. Hence, the only triple $(a,b,c)$ which further satisfies the condition that $a$, $b$, and $c$ are pairwise distinct is $(a,b,c)=(1,2,3)$.

Hint: since $$a|a+b+c\b|a+b+c\c|a+b+c$$therefore $$abc|a+b+c$$since $a,b,c$ are co-primes. Can you finish now?