Solution for a 5 variables diophantine equation-system (2 unknown independent variables, one given parameter)?

Question

In continuing/generalizing an earlier question(this and this) I arrived at the following problem on positive integers.

Assume $Q>0$ as given constant and either $(S,T) \ge 1$ as primary solutions are searched from which $(h,i) \ge 1$ can then be derived or other way round.
(If it helps we can moreover assume $S \le T$ resp. $h \le i$)

I can formulate my problem in the following diophantine equation-system: $$ \begin{array}{} QS &+ (Q-1)&=&hT \\ QT &+ (Q-1) &=&iS \end{array} \tag 1$$ Of course this can be rewritten as matrix-expression $$ \begin{bmatrix} h & -Q \\ -Q & i\end{bmatrix} \cdot \begin{bmatrix} T \\ S\end{bmatrix} = (Q-1)\begin{bmatrix} 1 \\ 1\end{bmatrix} \tag 2$$ Leftmultiply with inverse of the left matrix gives $$\begin{bmatrix} T \\ S\end{bmatrix} = {Q-1\over hi - Q^2} \cdot \begin{bmatrix} i & Q \\ Q & h\end{bmatrix} \cdot \begin{bmatrix} 1 \\ 1\end{bmatrix}= {Q-1\over hi - Q^2} \cdot \begin{bmatrix} Q+i \\ Q + h\end{bmatrix} \tag 3$$

But further fiddling towards a somehow parametrized version led me nowhere (actually: it led me into circles...) .

My first ansatz was assuming some $h$, for instance $h=1$ and then looking for the set of divisors available for $S$ from the first equation, but still I got no usable algebraic expression to improve this to indeterminate $h$.

Using an example with $Q=9$ I got a set of solutions using brute force by just checking $1\le h \le 17$ and $i$ from the lists of occuring possible divisors.

Q: is there any procedure which allows to avoid the brute-search (either in $(h,i)$ or in $(S,T)$)?

---

Example-solutions for $Q=9$

Searched by varying $(h,i)$ I give $s=S+1$ and $t=T+1$:

  s      t     h      i      S     T
 -------------------------------------
  2      2     17     17      1      1
  3      3     13     13      2      2
  5      5     11     11      4      4
  9      9     10     10      8      8
 29     53      5     17     28     52
  9     17      5     19      8     16
 89    401      2     41     88    400
  9     41      2     46      8     40
 81    729      1     82     80    728
 41    369      1     83     40    368
 21    189      1     85     20    188
 17    153      1     86     16    152
 11     99      1     89     10     98
  9     81      1     91      8     80
  6     54      1     97      5     53
  5     45      1    101      4     44
  3     27      1    121      2     26
  2     18      1    161      1     17

There are some patterns visible, but nothing discernable that would give me a hint towards some algebraic expression.

Answer 1

I have not yet a full algebraical answer, perhaps there is none.

Update
I found now a procedere where I can reduce the "brute-force" part to only apply to test all $h$ from the interval $1 \le h \lt Q$. With that the formulae given in the question allow then to determine possible solution for $i$,$s$,$t$ from sets of divisors (which is a massive reduction of the search space for $i$ compared with a "blind" search up to $Q^2-1$) .

It seems, that no better solution is possible.

So I can close the case here, and only shall try to formulate some more symmetric expression for the set of possible results as I'll perhaps find them.

A Pari/GP-routine which searches the list of solutions.

• First we test with $h=1$. From this a very simple expression for the set of solutions occur.
• Next we test $h=2 \ldots Q-1$. The same formula as before, but no reduction to such a simplicity. We need to determine for each $h$ one set of possible solutions for $i$ by the divisors of some expression dependend on $h$ and $Q$.
• Last we test $h=Q+1 \ldots 2Q-1$ . The formula shows, that $i=h$ is required and from this also $S=T$ is required. We get a small subset of a set of divisors as possible solutions for $h (=i)$

Pari/GP-code:

{doc_hi(Q=17)=my(QQ=Q^2,lhs,lhs1,di,idi,S,T,s,t,i,list,il);
 list=vectorv(10000);il=0;   \\ takes the list of results
 for(h=1,Q-1, 
     lhs=(Q+h)*(Q-1);
     di=divisors( lhs );   
     for(idi=1,#di, 
              S=di[#di+1-idi];s=S+1; \\ get S from the set of divisors
                  lhs1= di[idi]+QQ;
                  if(lhs1 % h, next()); \\ if integer division is not possible, skip
              i= lhs1 \ h;
              if( (Q*s-1) % h,next()); \\ if integer division is not possible, skip
              T= (Q*s-1)/h; t=T+1;
              il++;list[il]=[h,i,t,s,T,S];
  );  );          
  \\ Now check cases for h=i>Q, S=T
  di=divisors(Q-1); 
  for(idi=1,#di, S=T=di[#di+1-idi];s=S+1;t=T+1;
            h=i=Q+di[idi]; 
            il++; list[il]=[h,i,t,s,T,S];
     );
  list=VE(list,il); \\ shorten vector to il entries
  return(Mat(list));}

Results:

   > doc_hi(17)   \\ list results for Q=17
   h   i    t     s    T     S
   ----------------------------------------
   1  290  4913  289  4912  288
   1  291  2465  145  2464  144
   1  292  1649   97  1648   96
   1  293  1241   73  1240   72
   1  295   833   49   832   48
   1  297   629   37   628   36
   1  298   561   33   560   32
   1  301   425   25   424   24
   1  305   323   19   322   18
   1  307   289   17   288   16
   1  313   221   13   220   12
   1  321   170   10   169    9
   1  325   153    9   152    8
   1  337   119    7   118    6
   1  361    85    5    84    4
   1  385    68    4    67    3
   1  433    51    3    50    2
   1  577    34    2    33    1
   ---
   2  145  2593  305  2592  304
   2  154   145   17   144   16
   ---
   3   97   913  161   912  160
   3   98   369   65   368   64
   3   99   233   41   232   40
   3  103    97   17    96   16
   3  107    63   11    62   10
   3  123    29    5    28    4
   3  203    12    2    11    1
   ---
   4   73   481  113   480  112
   4   74   209   49   208   48
   ---
   5   58  1201  353  1200  352
   5   60   113   33   112   32
   5   61    79   23    78   22
   5   93    11    3    10    2
   ---
   6   52    49   17    48   16
   ---
   7   43    81   33    80   32
   7   55    13    5    12    4
   ---
   9   33   101   53   100   52
   9   35    33   17    32   16
   ---
  10   29   737  433   736  432
   ---
  11   27    89   57    88   56
  11   67     4    2     3    1
   ---
  13   23    65   49    64   48
   ---
   --- cases h=i >Q
   h   i    t     s    T     S
  ---------------------------------
  18   18    17   17    16   16
  19   19     9    9     8    8
  21   21     5    5     4    4
  25   25     3    3     2    2
  33   33     2    2     1    1

---

Old incomplete solution

(...)
But I can determine bounds for $h$ and $i$ assuming $h\le i$ and from this a search-space for $h$ and from assumed $h$ a set of values for $i$ and from this for $(S,T)$ resp $(s,t)=(S+1,T+1)$.

Assume $h=1$

We get by eq (1) $$ \begin{array} {} QS + (Q-1) &= 1 \cdot T &\qquad \qquad (1.1)\\ QT + (Q-1) &= i \cdot S &\qquad \qquad (1.2)\\ \end{array} \tag 1$$ Inserting $T$ in eq. (1.2) $$ \begin{array} {} \\ Q(QS + (Q-1)) + (Q-1) &= i \cdot S \\ Q^2S + (Q+1)(Q-1) &= i \cdot S \\ Q^2 + (Q^2-1)/S &= i \\ \end{array} \tag {2.1}$$ so $$ \begin{array} {} &S &\in \text{divisors}(Q^2-1) \\ &i &\in Q^2 + \text{divisors}(Q^2-1) \\ &\implies &(Q^2 +1) \le i \le (2Q^2-1) \end{array} \tag {2.2}$$

Assume $h=i$

We get by eq (1) $$ \begin{array} {} QS + (Q-1) &= i \cdot T &\qquad \qquad (3.1)\\ QT + (Q-1) &= i \cdot S &\qquad \qquad (3.2)\\ \end{array} \tag 3$$ Subtracting eq (3.1)- eq. (3.2) $$ \begin{array} {} \\ &Q(S -T) &= -i \cdot (S-T) \\ \text{if } S \ne T &Q &= -i & \text{impossible by problem definition}\\ \implies &S&=T &\text{required}\\ \end{array} \tag {4}$$ so $$ \begin{array} {} QS + (Q-1) &= i \cdot S &\qquad \qquad (5.1)\\ (Q-1) &= (i-Q) \cdot S &\qquad \qquad (5.2)\\ {Q-1 \over S} +Q&= i&\qquad \qquad (5.3)\\ \end{array} \tag 5$$ finally $$ \begin{array} {} &S(=T) &\in \text{divisors}(Q-1) \\ &i(=h) &\in Q + \text{divisors}(Q-1) \\ &\implies &(Q +1) \le i (=h) \le (2Q-1) \end{array} \tag {6}$$

Lower and upper bounds for $h$ and $i$

Since because of symmetry we can demand that $S \le T$ (or $h \le i$) we can determine the search space for $h$ as $$ 1 \le h \le 2Q-1 \tag 7$$ and for each $h$ find solution sets for $i$ from the sets of divisors of some formula depending on $h$.

This is not yet a true algebraical formula, but at least it is a strong reduction of search space for $h$ and then for $i$. One of the most interesting properties is, that for a given $Q$ the cardinality of the set of solutions is finite.