## 3 x 3 magic square w/ entries are distinct integer squares

$A^2$   …..   $B^2$   …..   $C^2$
$D^2$   …..   $E^2$   …..   $F^2$
$G^2$   …..   $H^2$   …..   $I^2$

is a magic square,

Can you find a number   $m$   such that

$A^2 + B^2 + C^2$
$= \; D^2 + E^2 + F^2$
$= \; G^2 + H^2 + I^2$
$= \; A^2 + D^2 + G^2$
$= \; B^2 + E^2 + H^2$
$= \; C^2 + F^2 + I^2$
$= \; A^2 + E^2 + I^2$
$= \; C^2 + E^2 + G^2 \; = \; m$

——————————————–

$A \; ..... \; B \; ..... \; C$
$D \; ..... \; E \; ..... \; F$
$G \; ..... \; H \; ..... \; I$

$A+B+C = 3 \,E \; ..... \; D+E+F = 3 \,E \; ..... \; G+H+I = 3 \,E \; ..... \; A+E+I = 3 \,E$
$A+D+G = 3 \,E \; ..... \; B+E+H = 3 \,E \; ..... \; C+F+I = 3 \,E \; ..... \; C+E+G = 3 \,E$

$A+I \; = \; B+H \; = \; C+G \; = \; D+F \; = \; 2 \,E$

$A+I \; = \; 2 \,E$
$(A-E) \; + \; (I-E) \; = \; 0$
$(A-E) \; = \; (E-I)$

If we define   $n = A-E$   and   $m = C-E$   the magic square contains the terms

$E+n \; .......... \; E+m$
$E$
$E-m \; .......... \; E-n$

From this the remaining terms follow,   due to the requirement that each row and column sum to   $3 \,E$

$E+n$                  $E-n-m$                  $E+m$
$E-n+m$                  $E$                          $E+n-m$
$E-m$                  $E+n+m$                  $E-n$

Imagine a magic square comprised entirely of square integers.

$a^2$     $b^2$     $c^2$
$d^2$     $e^2$     $f^2$
$g^2$     $h^2$     $i^2$

$a^2 = E+n$              $b^2 = E-n-m$             $c^2 = E+m$
$d^2 = E-n+m$              $e^2 = E$                     $f^2 = E+n-m$
$g^2 = E-m$              $h^2 = E+n+m$             $i^2 = E-n$

$a^2 + i^2 \; = \; b^2 + h^2 \; = \; c^2 + g^2 \; = \; f^2 + d^2 \; = \; 2 \,e^2$

$a^2 \, i^2 \; = \; (E+n) \,(E-n) \; = \; E^2 \; - \; n^2$
$a^2 \, i^2 \; + \; n^2 \; = \; E^2 \; = \; e^4$

$b^2 \, h^2 \; = \; (E-n-m) \,(E+n+m) \; = \; E^2 \; - \; (n+m)^2$
$b^2 \, h^2 \; + \; (n+m)^2 \; = \; E^2 \; = \; e^4$

$c^2 \, g^2 \; = \; (E+m) \,(E-m) \; = \; E^2 \; - \; m^2$
$c^2 \, g^2 \; + \; m^2 \; = \; E^2 \; = \; e^4$

$f^2 \, d^2 \; = \; (E+n-m) \,(E-n+m) \; = \; E^2 \; - \; (n-m)^2$
$f^2 \, d^2 \; + \; (n-m)^2 \; = \; E^2 \; = \; e^4$

math grad - Interest: Number theory
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### 4 Responses to 3 x 3 magic square w/ entries are distinct integer squares

1. Paul says:

That would be the holy grail of magic squares of squares if we could find that, if one exists (m), it is estimated to be over 15 digits in length.

P.

• benvitalis says:

For the smallest magic square,   $E$   must be odd, and all other squares are odd.   Since we want the square entries to be distinct, we need to choose   $E$   so that   $2 \, E^2$   can be expressed in at least four distinct ways as the sum of two squares

• Paul says:

It has been shown that the magic square of squares has the form

a² b² c² x+y, x-y-z, x+z
d² e² f² = x-y+z, x, x+y-z
g² h² i² x-z, x+y+z, x-y

and that x*y*z*(y+z)*(y-z) has to be divisible by 1546645545467664981281303961600. which is all primes <=67 except 59.

2. benvitalis says:

I’ll revisit the problem at a later time