Magic square of "almost"-squares
问题内容
I'm interested in a particular relaxation of the magic square of squares problem, where the entries are of the form $a^2\pm\varepsilon$ for some $\varepsilon$. The original problem is when $\varepsilon=0$.
By "correcting" the Parker square, there is a solution when $\varepsilon = 528$: $$\begin{pmatrix} 29^2 + \varepsilon & 1^2 -\varepsilon & 47^2-\varepsilon\\ 41^2-\varepsilon & 37^2 - \varepsilon & 1 + \varepsilon\\ 23 ^ 2-\varepsilon & 41^2 + \varepsilon & 29^2 - \varepsilon \end{pmatrix}=\begin{pmatrix} 1369 & -527 & 1681\\ 1153 & 841 & 529\\ 1 & 2209 & 313 \end{pmatrix},$$ where the magic sum is $2523$.
Question: how small can we make $\varepsilon$?
回答 (1)
Of course this may be not optimal. I continue search this, and will update when I found smaller $\epsilon$.
$\epsilon = 66$
- In this case, sum of row, column, diagonal are all $1677$. $$ \begin{pmatrix} 25^2+\epsilon & 5^2-\epsilon & 31^2+\epsilon\\ 31^2-\epsilon & 25^2-\epsilon & 17^2-\epsilon\\ 5^2+\epsilon & 35^2-\epsilon & 19^2+\epsilon \end{pmatrix} = \begin{pmatrix} 691 & -41 & 1027 \\ 895 & 559 & 223\\ 91 & 1159 & 427 \end{pmatrix} $$
$\epsilon = 444$
- In this case, sum of row, column, diagonal are all $2775$. $$ \begin{pmatrix} 29^2+\epsilon & 13^2-\epsilon & 47^2-\epsilon\\ 31^2+\epsilon & 37^2-\epsilon & 1^2+\epsilon\\ 23^2-\epsilon & 41^2+\epsilon & 11^2+\epsilon \end{pmatrix} = \begin{pmatrix} 1285 & -275 & 1765 \\ 1405 & 925 & 445\\ 85 & 2125 & 565 \end{pmatrix} $$