Unusual equation(s) obtained at an attempt to solve an open problem

Question

While trying to solve an open problem I arrived at the following equation:

$$56l^3+60l^2+10l+6=6\sum_{j=1}^{l+1}a_jb_j$$ , where $a_j,b_j$ are constrained by $1<a_j,b_j<4l$, for $j=1,...,l+1$

$a_j,b_j$ can be more precisely constrained but I will not go now into details because even this looks complicated enough.

Now, further constraints: $a_j$ and $b_j$ are also all odd and we have $a_j \neq b_j$ and we have that all $a_ib_i$ are different and we also must have $ 4 | a_jb_j - a_ib_i$ and we must also have this extremely strong divisibility condition $ (i-j) | a_ib_i-a_jb_j$.

Here we, in fact, have an infinite number of equations, because we get for every natural $l \in \mathbb N$ one non-linear Diophantine equation.

Honestly, I do not know who to consult when it comes to this problem.

What to do?

Answer 1

Subtract the equation for $l-1$ from the equation for $l$ and divide by 6 to get $a_{l+1}b_{l+1}=28l^2-8l+1<(4l)^2=16l^2$, so $12l^2-8l+1=(2l-1)(6l-1)<0$, i.e. $1/6<l<1/2$. But since $l$ is an integer, the problem has no solutions.

Update: Apparently, now $a_i$’s and $b_i$’s may be different for different $l$, so this answer has been rendered obsolete.

Answer 2

I have no idea whether the following description suffices for your purposes.

Let $n=\frac{56l^3+60l^2+10l+6}6$; note that $n$ is an integer.

Take any decomposition $n=m_1+...+m_{l+1}$ of $n$ into a sum of $l+1$ numbers which satisfy $m_i<16l^2$, $m_i\equiv m_j\mod4$, are all odd but neither primes nor squares of primes.

Choose any proper divisors $a_i|m_i$, $i=1,...,l+1$ with $a_i^2\ne m_i$, $m_i/4l<a_i<4l$. Then $(a_1,...,a_{l+1})$, $(b_1,...,b_{l+1})=(\frac{m_1}{a_1},...,\frac{m_{l+1}}{a_{l+1}})$ is a solution, and every solution can be obtained in this way.

There are no solutions for $l=2,3$; for $l=4$ there are, up to reorderings, 5 solutions:

(7,11,11,11,11),(15,15,15,15,15)
(7,11,11,11,15),(15,15,15,15,11)
(7,11,11,15,15),(15,15,15,11,11)
(7,11,15,15,15),(15,15,11,11,11)
(7,15,15,15,15),(15,11,11,11,11)

Update: with the last divisibility condition, there seem to be no solutions for $l=5,6,7,8$