Solve $a+b=\gcd(a^3,b^3),\;b+c=\gcd(b^3,c^3),\;c+a=\gcd(c^3,a^3)$ for positive integers $a,b,c$

Question

Solve $$\begin{cases}a+b=\gcd(a^3,b^3)\\ b+c=\gcd(b^3,c^3)\\ c+a=\gcd(c^3,a^3)\end{cases}$$ for positive integers $a,b,c$.

We can rewrite as: $$\begin{cases}a+b=\gcd(a,b)^3\\ b+c=\gcd(b,c)^3\\ c+a=\gcd(c,a)^3\end{cases}$$

I ran a program that checked $1\le a\le b\le c \le 500$, and it didn’t find any solutions. @RandomGuy checked all the numbers up to $200000$ and didn’t find any solutions, too (see in the comments). Probably there aren’t any.

Progress

I can show that $\color{red}{\gcd(a,b,c)=1}$. Indeed, let $d=\gcd(a,b,c)$, $a_1=ad$, $b_1=bd$, $c_1=cd$. Note that $\gcd(a,b)=\gcd(da_1,db_1)=d\cdot\gcd(a_1,b_1)$. Our system then becomes:

$$\begin{cases}a_1+b_1=d^2\cdot\gcd(a_1,b_1)^3\\ b_1+c_1=d^2\cdot\gcd(b_1,c_1)^3\\ c_1+a_1=d^2\cdot\gcd(c_1,a_1)^3\end{cases}\tag1$$

Note that $a_1+b_1+ c_1$ and $d$ are coprime. If $p\mid a_1+b_1+ c_1$ and $p\mid d$ then $p\mid d^2\cdot\gcd(a_1,b_1)^3 = a_1+b_1$. But then also $p\mid (a_1+b_1+ c_1)-(a_1+b_1)=c_1$. As well as $p\mid a_1$ and $p\mid b_1$. But $\gcd(a_1,b_1,c_1)=1$ by definition of $a_1$, $b_1$, $c_1$.

Now add all three equations of $(1)$: $$2(a_1+b_1+ c_1)= d^2\cdot\left(\gcd(a_1,b_1)^3+ \gcd(b_1,c_1)^3 +\gcd(c_1,a_1)^3\right).$$ Since $a_1+b_1+ c_1$ and $d$ are coprime, we have $d^2\mid 2.$ So $$d=1.$$

Motivation

This system of equations came from me experimenting with a code to bruteforce different equations. A single equation has a lot of solutions, while the system of three equations with three variables seemingly has none.

Here are all the solutions of $a+b=\gcd(a^2,b^2)$, where $1\le a\le b\le 500$. The three columns are $a$; $b$; $\gcd(a,b)$.

2 2 2
3 6 3
4 12 4
5 20 5
6 30 6
7 42 7
8 56 8
9 72 9
10 15 5
10 90 10
11 110 11
12 132 12
13 156 13
14 35 7
14 182 14
15 210 15
16 240 16
17 272 17
18 63 9
18 306 18
19 342 19
20 380 20
21 28 7
21 420 21
22 99 11
22 462 22
24 40 8
26 143 13
30 70 10
30 195 15
33 88 11
34 255 17
36 45 9
38 323 19
39 130 13
42 154 14
42 399 21
44 77 11
46 483 23
48 208 16
51 238 17
52 117 13
55 66 11
57 304 19
60 84 12
60 165 15
60 340 20
65 104 13
66 418 22
68 221 17
69 460 23
70 126 14
76 285 19
78 91 13
80 176 16
84 357 21
85 204 17
90 234 18
92 437 23
95 266 19
102 187 17
105 120 15
105 336 21
110 374 22
112 144 16
114 247 19
115 414 23
119 170 17
120 456 24
126 198 18
133 228 19
136 153 17
138 391 23
140 260 20
150 475 25
152 209 19
154 330 22
161 368 23
168 273 21
168 408 24
171 190 19
175 450 25
180 220 20
182 494 26
184 345 23
198 286 22
200 425 25
207 322 23
210 231 21
225 400 25
230 299 23
234 442 26
253 276 23
264 312 24
270 459 27
275 350 25
286 390 26
297 432 27
300 325 25
308 476 28
348 493 29
351 378 27
364 420 28
377 464 29
406 435 29
465 496 31

Here are all the solutions of $a+b=\gcd(a^3,b^3)$, where $1\le a\le b\le 500$. Interestingly, there are no solutions where $a$ and $b$ both are in $\{400,…,499\}$. Columns are $a$; $b$; $\gcd(a,b)$.

2 6 2
3 24 3
4 60 4
5 120 5
6 21 3
6 210 6
7 336 7
10 115 5
12 15 3
12 52 4
14 329 7
15 110 5
20 44 4
20 105 5
21 322 7
24 488 8
28 36 4
28 315 7
30 95 5
30 186 6
35 90 5
35 308 7
40 85 5
40 472 8
42 174 6
42 301 7
45 80 5
55 70 5
56 287 7
56 456 8
60 65 5
63 280 7
66 150 6
70 273 7
72 440 8
77 266 7
78 138 6
84 259 7
88 424 8
91 252 7
102 114 6
104 408 8
105 238 7
112 231 7
119 224 7
120 392 8
126 217 7
133 210 7
136 376 8
140 203 7
152 360 8
154 189 7
161 182 7
168 175 7
168 344 8
184 328 8
200 312 8
216 296 8
232 280 8
234 495 9
248 264 8
252 477 9
261 468 9
279 450 9
288 441 9
306 423 9
315 414 9
333 396 9
342 387 9
360 369 9

Answer 1

Adding all three equations we get $$\sum_{cyc}\gcd(a,b)^3=2(a+b+c)$$

Since the sum of gcds are even either all must be even or one of them must be even

Since atleast one $\gcd$ is even this guarantees 2 numbers are even

Case 1: Assume all the numbers are even then the system reduces to

$$ a_1+b_1=4\gcd(a_1,b_1)^3\\ a_1+c_1=4\gcd(a_1,c_1)^3\\ b_1+c_1=4\gcd(b_1,c_1)^3 $$

Solving for $a_1$ we get $$a_1=2(\gcd(a_1,b_1)^3+\gcd(a_1,c_1)^3-\gcd(b_1,c_1)^3)$$

And similar expressions for $b_1,c_1$ show that they are even too

Now we can create $a_2,b_2,c_2$ in a similar fashion and keep going till infinity which would mean the power of 2 in $a,b,c$ is $\infty$ this is absurd so no numbers satisfy for this case

Case 2: assume a,b are even and c is odd

Edit: No solutions till 200,000 by brute force

Now this case I'm not entirely sure how to prove so I'll leave with just the proof for case 1

Answer 2

I notice that $\gcd(a,b)$ is much smaller than $a$ and $b$ so I'll try to find a way to iterate through values of gcd in stead.

Let

$$u = \gcd\left(a,b\right), v = \gcd\left(b,c\right), w = \gcd\left(a,c\right)$$ then $$ a = a_b u = a_c w$$ $$ b = b_a u = b_c v$$ $$ c = c_a w = c_b v$$ The system can be reduced as $$ \begin{cases} \begin{align} a_b + b_a &= u^2\\ b_c + c_b &= v^2\\ c_a +a_c &= w^2 \end{align} \end{cases} $$ Since $\gcd(a,b,c) = 1$ as shown by Aig, $u,v,w$ are pairwise coprime. This means $(u \mid a_c, b_c)$, $(v \mid b_a, c_a)$ and $(w \mid a_b, c_b)$. Notice that since $a_b u = a_c w$, then $a_b/w = a_c/u$ . Let $x = a_b/w = a_c/u$, $y = b_c/u = b_a/v$, $z = c_a/v = c_b/w$, then the system can be written as follows $$ \begin{cases} \begin{align} xw + yv &= u^2\\ yu + zw &= v^2\\ zv +xu &= w^2 \end{align} \end{cases} $$ Also since $1 = \gcd(a_b,b_a) = \gcd(xw, yv)$, $x$ and $y$ are coprime, similarly, $x,y,z$ are pairwise coprime. As mentioned by Random guy, at least one among $u, v, w$ is even. WLOG, assume $u$ is even. Then $a$ and $b$ are even while both $v$ and $w$ are odd, as well as $c$. Since $z\mid c$, $z$ is odd. Since $u^2$ is even and $v,w$ are odd, $x \equiv y \mod 2$. Because they are coprime, x and y must also be odd.

So in short, $u, v, w$ are pairwise coprime where one of them is even, $x, y, z$ are coprime odds that satisfy the given Diophante equation.