Solutions $a,b,c \mid 1+a+b+c $ over positive integers algebraically (avoiding brute force)?

Question

I arrived at the diophantine problem for $3$ positive integers that should maintain all 3 divisibilities jointly: $$ a | 1+a+b+c \\\ b | 1+a+b+c \\\ c | 1+a+b+c \tag 1 \\\ $$ I tried putting up a matrix expression introducing positive integer parameters $(i,j,k)$ $$\begin{array}{} & \\ \begin{bmatrix} -i&1&1 \\ 1&-j&1 \\ 1&1&-k \\ \end{bmatrix} & \cdot \begin{bmatrix} a\\b\\c \end{bmatrix} =\begin{bmatrix} -1\\-1\\-1 \end{bmatrix} \end{array} \tag 2 $$ but the fiddling with the occuring formulae using $(i,j,k)$ now is again not yet conclusive.

I didn't find another convincing ansatz towards a formula.

By brute force ($2\le a \le b \le c \le 120$) I found the following solutions (avoiding symmetries) for $[1,a,b,c]$

[1, a, b, c]
--------------
[1, 2, 2, 5]
[1, 2, 3, 6]
[1, 2, 6, 9]
[1, 3, 4, 4]
[1, 3, 8, 12]
[1, 4, 5, 10]        
[1, 6, 14, 21]       

Searching using eq(2) with $1\le i \le j \le k \le 32$ I got

[i, j, k] ---> [1, a,   b, c]  rotated
-------------------------------
[1, 2, 6] ---> [1,21,14, 6]      R
[1, 2, 7] ---> [1,12, 8, 3]      R
[1, 2, 8] ---> [1, 9, 6, 2]      R
[1, 2,11] ---> [1, 6, 4, 1]
[1, 3, 4] ---> [1,10, 5, 4]      R
[1, 3, 5] ---> [1, 6, 3, 2]      R
[1, 3, 7] ---> [1, 4, 2, 1]
[1, 4, 4] ---> [1, 5, 2, 2]      R
[1, 5, 5] ---> [1, 3, 1, 1]
[2, 2, 3] ---> [1, 4, 4, 3]      R
[2, 2, 5] ---> [1, 2, 2, 1]
[3, 3, 3] ---> [1, 1, 1, 1]

which are simply rotated versions or trivials (having $a or b or c=1$) which are excluded in the above list.

I guess these are all possible solutions, but don't find the argument...

Q1: How can I approach that problem algebraically?
Q2: is the number of solutions finite or infinite?
Q3.1: if the set of solutions is infinite is there a parametrization?
Q3.2: if the set of solutions is finite, what is that set?

---

update In generalizing the problem towards [h,a,b,c] with $h \in \mathbb N^+$ I seem to get the full list of $14$ solutions (avoiding solutions with $\gcd()>1$ and rotations) which are

[h, a, b, c]         some interpretations 
--------------------------------------------
[1, 1, 1, 1]
[1, 1, 1, 3]
[1, 1, 2, 2]
[1, 1, 2, 4]
[1, 1, 4, 6]
[1, 2, 2, 5]          1+2+2=5:   5=5/1
[1, 2, 3, 6]        1+2=3:  3=1*3  6=2*3
[1, 2, 6, 9]        1+2=3:  6=2*3  9=3*3
[1, 3, 4, 4]        1+3=4:  4=1*4  4=1*4
[1, 3, 8, 12]       1+3=4:  8=2*4 12=3*4
[1, 4, 5, 10]       1+4=5:  5=1*5 10=2*5
[1, 6, 14, 21]      1+6=7: 14=2*7 21=3*7
[2, 3, 3, 4]          2+3+3=8:  4=8/2
[2, 3, 10, 15]      2+3=5: 10=2*5 15=3*5

A bit context: this is a detail-problem in an earlier question where I explore the general conditions in the Lehmer's totient problem. In the earlier question I've considered a three-variable $(R,S,T)$ diophantine system, and looked at solutions of the form $(R,S,T)=(R^1,R^a,R^b)$ Here I generalize it to $4$ variables and solutions of $(Q,R,S,T)=(Q^1,Q^a,Q^b,Q^c)$ resp. $(Q,R,S,T)=(Q^h,Q^a,Q^b,Q^c)$ ($\gcd(h,a,b,c)=1$) and determine the solutions in terms of $(1,a,b,c)$ resp. $(h,a,b,c)$. I'll later generalize to more variables but first I want to get some grip about the general limitations and whereabouts - at best in a form which support later generalizations...

Answer 1

I might have found an elegant solution, easily generalizable to more variables (which is why I didn't stop with @aqua's nice solution).

This solution is found by the even more general ansatz using $(q,a,b,c)$ instead of $(1,a,b,c)$ as formulated in my OP.

I start with the matrix-formula $$ \begin{matrix} \left [ \begin{smallmatrix} 1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1 \end{smallmatrix} \right ] & * & \left [ \begin{smallmatrix} q\\a\\b\\c \end{smallmatrix} \right ] &=& \left [ \begin{smallmatrix} h&.&.&.\\.&i&.&.\\.&.&j&.\\.&.&.&k \end{smallmatrix} \right ] &*&\left [ \begin{smallmatrix} q\\a\\b\\c \end{smallmatrix} \right ] \end{matrix} \tag 1$$ We have an additional restriction by the demand that $q \le a\le b\le c $, which I encoded into the formula writing $$ q=q_1, a=q_1+a_1, b=q_1+a_1+b_1, c= q_1+a_1+b_1+c_1 \\ \qquad \text{now with } q_1\ge 1, \text{ and } a_1,b_1,c_1 \ge 0 $$ Rearranging eq(1) for the new variables $q_1,a_1,b_1,c_1$ and accounting the rhs into the lhs gives a form of an eigenvector-problem $M_1 \cdot A = 0$ as $$ \begin{matrix} \left [ \begin{smallmatrix} -h+4&3&2&1\\ -i+4&-i+3&2&1\\ -j+4&-j+3&-j+2&1\\ -k+4&-k+3&-k+2&-k+1 \end{smallmatrix} \right ]&*& \left [ \begin{smallmatrix} q_1\\a_1\\b_1\\c_1 \end{smallmatrix} \right ]&=& \left [ \begin{smallmatrix} 0\\0\\0\\0 \end{smallmatrix} \right ] \end{matrix} \tag 2$$ Such a system can nontrivially only be solved when the determinant of the left matrix multiplicator is zero, so when $ \qquad \text{matdet}(M_1) = 0 \qquad$.

The determinant of $M_1$ can easily be computed, it gives the expression with the unknowns $(h,i,j,k)$ $$ \text{matdet}(M_1) = hijk -( hij + ijk + jkh + khi) \tag 3 $$ and demanding this to be zero gives the elsewhere well known problem in integers $$ \text{matdet}(M_1) = 0 \implies 1 = \frac1h + \frac1i + \frac1j + \frac1k \tag 4 $$ This can be searched with a small search-space (or be dealt analoguously as in @aqua's answer) giving the known set of $14$ solutions

[h, i, j,  k] h=2
---------------------  
[2, 3, 7, 42]
[2, 3, 8, 24]
[2, 3, 9, 18]
[2, 3, 10, 15]
[2, 3, 12, 12]
[2, 4, 5, 20]
[2, 4, 6, 12]
[2, 4, 8, 8]
[2, 5, 5, 10]
[2, 6, 6, 6]

[h, i, j,  k] h=3
---------------------  
[3, 3, 4, 12]
[3, 3, 6, 6]
[3, 4, 4, 6]

[h, i, j,  k] h=4
---------------------  
[4, 4, 4, 4]

Fixing $q_1=1$ allows to compute the solutions for $a_1,b_1,c_1$ from this (and then for $(q,a,b,c)$ from this. Sometimes $a_1,b_1,c_1$ becomes fractional- but then we can normalize by multiplication by the $\operatorname{lcm}()$ of the common denominator bringing $q$ to a value greater than $1$.

The problem of finding $1=1/h+1/i+1/j+1/k$ in integers has not yet an algebraic solution (due to mathworld,"egyptian numbers" and "egyptian fraction") and must essentially still be solved by a search-routine, but the search-space for the parameter $(h,i,j,k)$ is much smaller than that for $(q,a,b,c)$ directly.

I got the following result:

[h, i, j, k]    [q, a,  b,  c]  error-check
-------------------------------------------
[2, 3, 7, 42]   [1, 6, 14, 21] ok
[2, 3, 8, 24]   [1, 3, 8, 12]  ok
[2, 3, 9, 18]   [1, 2, 6, 9]   ok
[2, 3, 10, 15]  [2, 3, 10, 15] ok
[2, 3, 12, 12]  [1, 1, 4, 6]   ok
[2, 4, 5, 20]   [1, 4, 5, 10]  ok
[2, 4, 6, 12]   [1, 2, 3, 6]   ok
[2, 4, 8, 8]    [1, 1, 2, 4]   ok
[2, 5, 5, 10]   [1, 2, 2, 5]   ok
[2, 6, 6, 6]    [1, 1, 1, 3]   ok
[3, 3, 4, 12]   [1, 3, 4, 4]   ok
[3, 3, 6, 6]    [1, 1, 2, 2]   ok
[3, 4, 4, 6]    [2, 3, 3, 4]   ok
[4, 4, 4, 4]    [1, 1, 1, 1]   ok                 

which is, besides of rotation, the same result as I got in my OP (by a much wider search space on $(q,a,b,c)$).

---

Generalization It seems -after short checking- that the formula eq(4) looks analoguously generalized when we generalize the number of variables ($3$ or $5$), but I've to look at this deeper first. For the case of $5$ variables, prepending $g$ to the set I get for the determinant $$ \text{matdet}(M5) = -ghijk + (ghij+hijk+ijkg+jkgh+kghi) \tag 5 $$ determining the solutionspace $$ 1 = \frac1g+\frac1h+\frac1i+\frac1j+\frac1k \tag 6 $$ and I'm sure this extends easily for the $6$- or more variables problem.

Answer 2

Because of symmetry, you can assume that $a\leq b\leq c$, then we have $$ck= 1+a+b+c \implies k\leq 4$$

• If $\boxed{k=1}$ we have $c=1+a+b+c$ a contradiction

---

• If $\boxed{k=2}$ we have $2c=1+a+b+c$ so $c= 1+a+b$. Then $b\mid 2+2a+2b$ so $b\mid 2+2a$ so $mb = 2a+2\leq 4b\implies m\leq 4$ and now is easy to check all 4 subscases...

  • If $m=1$ we get $b=2a+2\implies a\mid 2+2a+4a+4\implies a\mid 6$...
  • If $m=2$ we get $b=a+1\implies a\mid 2+2a+2a+2\implies a\mid 4$...
  • If $m=3$ we get $3b=2a+2\leq 2b+2\implies b\leq 2$ so $b=2$ and $a=2$.
  • If $m=4$ we get $4b=2a+2 \leq 2b+2\implies b=1$

---

• If $\boxed{k=3}$ we have $3c=1+a+b+c$ so $2c=1+a+b$.

  • If $b\leq c-1$ then $2c\leq 1+2c-2 = 2c-1$ a contradiction.

  • If $b= c$ then $c=1+a$ and now we have $a\mid 3+3a \implies a\mid 3$ so $a=1$ and $b=c=2$ or $a=3$ and $b=c=4$.

---

• If $\boxed{k=4}$ we have $4c = 1+a+b+c$ so $a=b=c=1$.

Answer 3

"OP" Equation is equivalent to :

$ap=bq=cr=m$ ----(1)

where, $m=(a+b+c+1)$

$(p,q,r)$ are factor's of $'m'$

equation (1) has parametric solution:

$(p,q,r)=(t,2t,3t)$

$(a,b,c)=(6w,3w,2w)$

Where, $w=[1/(6t-11)]$

for, $t=7$ we get w=(1/31) &,

$(p,q,r)=(7,14,21)$

$(a,b,c)=[(6/31),(3/31),(2/31)]$

$m=(42/31)$