Finding the answer
It's very easy to strongly conjecture the answer if we don't need proof. (Then I include a full proof below.)
First, observe that if $(a,b,c)$ is a solution then $(ka, kb, kc)$ is also a solution for any $k \in \mathbb{N}$, so we can assume WOLOG that $\gcd(a,b,c) = 1$. Second, just check all smallish cases using a computer. I used Python:
from math import gcd
M = 300 # We'll check all a,b,c <= this cutoff.
for a in range(1, M+1):
for b in range(a, M+1):
for c in range(b, M+1):
if gcd(gcd(a,b),c) == 1 and (gcd(a,b) + gcd(b,c) + gcd(a,c)) * 2 == a+b+c:
print(a,b,c)
The output of this program makes it look very probably that the primitive solutions are: $(1,2,3)$, $(1,3,6)$, plus $(4,a,a)$ for any $a$ with $\gcd(4,a)=1$. (Reorderings of these terms are also solutions.)
Proving the answer
This gets pretty messy! I'll break into the following cases:
Case 1: $a = b$
We assumed $\gcd(a,b,c) = 1$, so $\gcd(a,c) = 1$. That means we need $2a + c = 2(2 + a) = 4 + 2a$, so $c = 4$. In other words the solution must take the form $(4,a,a)$ with $\gcd(4,a) = 1$. Any tuple of that form works, which gives infinitely many solutions.
Case 2: $a \mid b$ and $a \ne b$
Let $b = ka$. Define $g = \gcd(k,c)$, $c = gz$, $k = gy$. We have $\gcd(a,b) = a$, $\gcd(a,c) = 1$ (since otherwise $\gcd(a,b,c)$ would be bigger than 1) and $\gcd(b,c) = g$. The equation becomes
$$\begin{align}
2(a + g + 1) &= a + gya + gz \\
2g+2 &= (gy-1)a + gz \\
g(ay + z - 2) &= a + 2 \\
a(gy - 1) &= 2 + 2g - zg
\end{align}$$
We can use those last two lines to check some rapid-fire subcases:
- If $z \ge 5$ then $g(ay+z-2) \ge g(ay+3) \ge a+3$, so there's no hope. Thus $z \le 4$.
- If $y \ge 6$ then $g(ay+z-2) = g(2a + z + (y-2)a-2) \ge g(2a + z + 2) > 2a+2$. Thus we must have $y \le 5$.
- We must have $ay+z-2 > 0$, and $ay+z-2 \ge a-1$. The largest possible ratio of $\frac{a+2}{\max(1, a-1)}$ is 3, so we must have $g \le 3$.
- Using the observations above, we have $2 + 2g - zg \le 2 + 6 - 1 = 7$. Using the assumption $a \ne b$ and thus $gy > 1$, we have $a(gy-1) \ge a$. Therefore we must have $a \le 7$.
It's possible to be a little more careful and prove tighter bounds, but anyway the above observations combine to tell us $a \le 7$, $g \le 3$, $y \le 5$, $z \le 4$. That leaves only 420 cases to check, which takes no time at all in Python or any other language you choose. The only solutions in this category are $(a,b,c) = (1,2,3), (1,3,2), (1,4,1), (3,6,1)$. Discarding $(1,3,2)$ which is just a rearrangement and $(1,4,1)$ which is of the form $(4,a,a)$ from the case above, the only interesting new solutions here are $(1,2,3)$ and $(1,3,6)$.
Here's the Python code I used to check the 420 cases:
from math import gcd
for a in range(1, 8):
for g in range(1, 2):
for y in range(1, 6):
for z in range(1, 5):
b = g*y*a
c = g*z
if gcd(gcd(a,b),c) != 1: continue
if a == b: continue
if g*(a*y + z - 2) == a + 2: print(a,b,c)
Case 3: No two of $a,b,c$ divide each other
Let $a \le b \le c$ and note $\gcd(a,b) \le \frac{a}{2}$, $\gcd(b,c) \le \frac{b}{2}$, and $\gcd(a,c) \le \frac{a}{2}$. Then
$$\begin{align}
2(\gcd(a,b) + \gcd(b,c) + \gcd(a,c)) &\le 2\left( \frac a 2 + \frac b 2 + \frac a 2 \right) \\
&= 2a+b < a+b+c.
\end{align}$$
Thus there can't be any solutions in this case.
Finale: Combining the results from the cases
The full set of solutions $(a,b,c)$ exactly contains:
- $(1,2,3)$ and $(1,3,6)$
- Any tuple $(4,a,a)$ where $\gcd(4,a) = 1$
- Any rearranged and/or scaled version of one of the above options
(Of course, we were pretty sure this would be the result after 1 minute of coding in Python at the very top of this answer. But now we've actually proved it!)
