Theorem 1, by Lee Morgenstern, November 2006.
In a 3x3 magic square of distinct squares, the smallest square cannot be 1.

Lemma 1
Given a sequence of positive integers,
   s[1], s[2], ...
   s[j] > 2s[j-1] for all j > 1,
there is no solution to
   s[k] = s[i] + s[j]
with distinct i,j,k.

Lemma 2
All possible solutions to the equation
   2r^2 - p^2 = 1
in positive integers are produced

in increasing order of magnitude by the recursion
   p[n] = 3p[n-1] + 4r[n-1]
   r[n] = 3r[n-1] + 2p[n-1]
   p[1] = r[1] = 1.

Lemma 3
The entries of any 3x3 magic square of positive distinct integers can be represented by expressions of the three positive terms,
   a,b,c, 0 < a < b,
in the arrangement
   c+b      c-(a+b)  c+a
   c-(b-a)  c        c+(b-a)
   c-a      c+(a+b)  c-b
where the smallest entry is

Proof of Theorem 1
From Lemma 3, the smallest entry is the starting value of the three arithmetic progressions of entries
   c-(a+b), c-b, c-(b-a)
   c-(a+b), c-a, c+(b-a)
   c-(a+b), c,   c+(a+b)
having step values a, b, (a+b), respectively.

So, in a 3x3 magic square of distinct squares with 1 as the smallest entry, there must be three sets of three squares in arithmetic progression, each starting with 1 and having distinct step values
   s[i], s[j], s[k]
   s[k] = s[i] + s[j].

 The three-square arithmetic progression
   1, r^2, p^2
with step value s is a solution to
   1 + s = r^2
   r^2 + s = p^2
   2r^2 - p^2 = 1
   s = r^2 - 1.

 Lemma 2 gives the solutions of
   2r^2 - p^2 = 1
for p, r and then from r we can compute s.
This produces the sequences
  p[n]: 1,  7,  41,   239, ...
  r[n]: 1,  5,  29,   169, ...
  s[n]: 0, 24, 840, 28560, ...

 From the recursion formula in Lemma 2, we see that each r value is more than 3 times larger than the previous r value.
Thus the s value is more than 9 times the value of the previous s value.
Therefore, from Lemma 1, there is no solution to
   s[k] = s[i] + s[j]
using three distinct positive values of s from the sequence.

 Proof of Lemma 1
Since all values are positive and in increasing order of magnitude, if
   s[k] = s[i] + s[j]
then we must have
   k > j, k > i
and thus
   s[k] > 2s[j].

 If also j > i, then
   s[j] > 2s[i]
   (1/2)s[j] > s[i]
and then adding s[j] to both sides,
   (3/2)s[j] > s[i] + s[j].

 Combining the above, we have
   s[k] > 2s[j] > (3/2)s[j] > s[i] + s[j].
   s[k] > s[i] + s[j]
for any distinct i,j,k, k > j > i.

Proof of Lemma 2
See any reference on the Pell Equation from Number Theory or Recreational Mathematics.

Proof of Lemma 3
See any reference on 3x3 magic squares.

 About Lemma 1 and Theorem 1

These are original from Lee Morgenstern.

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