Problem:
IBM Research - Ponder This - January 2019 monthly contest (which was closed few days ago) leads to the problem:
Find sets $A = \{a_1,a_2,\ldots,a_n\}$, $B = \{b_1,b_2,\ldots, b_m\}$ such that $a_i+b_j = s_{ij}^2$
for $n\ge 4, m\ge 4$ (where $a_i,b_j,s_{ij}\in\mathbb{Z}$, $0\le a_1 < a_2 < \ldots < a_n$, $0\le b_1 < b_2 < \ldots < b_m$).
And it was said that nothing is known about solution for the case $n=m=5$.
Related links:
Approach:
If focus on the case $n=m$, then w.l.o.g. $a_1 \le b_1$; and since any solution $(A,B) =(\{a_i\}, \{b_j\})$ can be written in "shifted" form $(A',B')=(\{a_i-c\}, \{b_j+c\})$, $-$ we'll focus on solutions with $a_1 = 0$.
Actually, it's enough to write $a_i$, $b_j$ in the form:
$a_i = x_i^2 - \omega^2$,
$b_j = y_j^2$,
$( x_1 = \omega$, $ y_1 = \omega$, $ \omega\ge 0)$, to find solution of the problem.
To keep symmetry, one can write key numbers of solution in "square" table \begin{array}{c|c|c|c|c|c|} \color{tan}{-\omega^2} & \color{blue}{\omega^2} & \color{blue}{y_2^2} & \color{blue}{y_3^2} & \color{blue}{\ldots} & \color{blue}{y_m^2} \\ \hline \color{red}{\omega^2} & \color{gray}{\omega^2} & \color{gray}{y_2^2} & \color{gray}{y_3^2} & \ldots & \color{gray}{y_m^2} \\ \hline \color{red}{x_2^2} & \color{gray}{x_2^2} & s_{22}^2 & s_{23}^2 & \ldots & s_{2m}^2 \\ \hline \color{red}{x_3^2} & \color{gray}{x_3^2} & s_{32}^2 & s_{33}^2 & \ldots & s_{3m}^2 \\ \hline \color{red}{\vdots} & \color{gray}{\vdots} & \vdots & \vdots & \ddots & \vdots \\ \hline \color{red}{x_n^2} & \color{gray}{x_n^2} & s_{n2}^2 & s_{n3}^2 & \ldots & s_{nm}^2 \\ \hline \end{array} where $$s_{ij}^2 = \color{red}{x_i^2} + \color{blue}{y_j^2} - \omega^2.\tag{1}$$
Example:
One of $4\times 4$ examples:
$A = \{0, \; 282^2-18^2, \; 477^2-18^2,\; 1122^2 - 18^2\},$
$B=\{18^2, 234^2, 346^2, 514^2\}$ with appropriate table:
\begin{array}{c|c|c|c|c|} \color{tan}{-18^2} & \color{blue}{18^2} & \color{blue}{234^2} & \color{blue}{346^3} & \color{blue}{514^2} \\ \hline \color{red}{18^2} & \color{gray}{18^2} & \color{gray}{234^2} & \color{gray}{346^2} & \color{gray}{514^2} \\ \hline \color{red}{282^2} & \color{gray}{282^2} & 366^2 & 446^2 & 586^2 \\ \hline \color{red}{477^2} & \color{gray}{477^2} & 531^2 & 589^2 & 701^2 \\ \hline \color{red}{1122^2} & \color{gray}{1122^2} & 1146^2 & 1174^2 & 1234^2 \\ \hline \end{array}
Efforts:
I was/am excited to find $A,B$ for $n=m=5$. Unfortunately, without (essential) success: best result I can reach is finding two $5\times5$ tables-"siblings", which have $24$ of $25$ square numbers inside:
\begin{array}{c|c|c|c|c|c|} \color{tan}{-360^2} & \color{blue}{360^2} & \color{blue}{19240^2} & \color{blue}{41535^3} & \color{blue}{79560^2} & \color{blue}{161928^2} \\ \hline \color{red}{360^2} & \color{gray}{360^2} & \color{gray}{19240^2} & \color{gray}{41535^2} & \color{gray}{79560^2} & \color{gray}{161928^2} \\ \hline \color{red}{24696^2} & \color{gray}{24696^2} & 31304^2 & 48321^2 & 83304^2 & 163800^2 \\ \hline \color{red}{39060^2} & \color{gray}{39060^2} & 43540^2 & 57015^2 & \color{orange}{7855347600} & 166572^2 \\ \hline \color{red}{81480^2} & \color{gray}{81480^2} & 83720^2 & 91455^2 & 113880^2 & 181272^2 \\ \hline \color{red}{131460^2} & \color{gray}{131460^2} & 132860^2 & 137865^2 & 153660^2 & 208572^2 \\ \hline \end{array}
and
\begin{array}{c|c|c|c|c|c|} \color{tan}{-360^2} & \color{blue}{360^2} & \color{blue}{19240^2} & \color{blue}{41535^3} & \color{blue}{51480^2} & \color{blue}{161928^2} \\ \hline \color{red}{360^2} & \color{gray}{360^2} & \color{gray}{19240^2} & \color{gray}{41535^2} & \color{gray}{51480^2} & \color{gray}{161928^2} \\ \hline \color{red}{24696^2} & \color{gray}{24696^2} & 31304^2 & 48321^2 & 57096^2 & 163800^2 \\ \hline \color{red}{39060^2} & \color{gray}{39060^2} & 43540^2 & 57015^2 & 64620^2 & 166572^2 \\ \hline \color{red}{81480^2} & \color{gray}{81480^2} & 83720^2 & 91455^2 & \color{orange}{9289051200} & 181272^2 \\ \hline \color{red}{131460^2} & \color{gray}{131460^2} & 132860^2 & 137865^2 & 141180^2 & 208572^2 \\ \hline \end{array}
Question: Is it possible to find solution for the case $5\times 5$? Does there exist one? Any ideas or estimations?