Impossibility of normal tetramagic square 25x25
by Lee Morgenstern, April 2013.
S1 = 7825
S2 = 3263025
S3 = 1530765625
S4 = 765994466145 = 1 (mod 16)
Since S4 = 1 (mod 16), there must be either 1 or 17 odd entries in a tetramagic series.
In the case of 1 odd entry and 24 even entries:
Let Ej = 2Fj be an even entry, j = 1 ... 24.
Let G = 2K+1 be the odd entry.
Let Pn = sum (Fj)^n, j = 1 ... 24, n = 1 ... 4.
S1 = sum (2Fj) + G =
2P1 + 2K + 1 = 7825
S2 = sum (2Fj)^2 + G^2 = 4P2 + 4K^2 + 4K + 1 = 3263025
S3 = sum (2Fj)^3 + G^3 = 8P3 + 8K^3 + 12K^2 + 6K + 1 = 1530765625
S4 = sum (2Fj)^4 + G^4 = 16P4 + 16K^4 + 32K^3 + 24K^2 + 8K + 1 = 765994466145
P2 + K(K+1) = 815756
K(K+1) must be even and 815756 is even, thus P2 is even.
Since P1,P2,P3,P4 are sums of powers of the same entries, they all must be even.
P1 + K = 3912.
P1 is even and 3912 is even, thus K is even.
Let K = 2L. From S3,
2P3 + 16L^3 + 12L^2 + 3L = 382691406 = 2 (mod 4)
Since P3 is even, 2P3 is a multiple of 4 as well as 16L^3, and 12L^2. thus L = 2 (mod 4).
Let L = 4M+2 and thus K = 8M+4 and G = 16M+9. Rewriting the equations in terms of M, we have
P1 + 8M = 3908 = 0 (mod 4)
P2 + 64M^2 + 72M = 815736 = 0 (mod 8)
P3 + 512M^3 + 864M^2 + 486M = 191345612 = 0 (mod 4)
P4 + 4096M^4 + 9216M^3 + 5728M^2 + 2916M = 47874653724 = 4 (mod 8)
P1 = 0 (mod 4)
P2 = 0 (mod 8)
P3 = 2 (mod 4) if M is odd
= 0 (mod 4) if M is even
P4 = 0 (mod 8) if M is odd
= 4 (mod 8) if M is even
An even cube = 0 (mod 4), thus P3 = sum of odd cubes (mod 4).
An odd entry = that entry cubed (mod 4).
Sum of even entries = (P1 - sum of odd entries) (mod
= (P1 - sum of odd cubes) (mod 4)
= (P1 - P3) (mod 4),
sum of even entries = 2 (mod 4) if M is odd,
= 0 (mod 4) if M is even.
An even 4th power = 0 (mod 8), thus P4 = sum of odd 4th powers (mod 8).
An odd entry squared = 4th power of that entry (mod 8).
Sum of even squares = (P2 - sum of odd squares) (mod
= (P2 - sum of odd 4th powers) (mod 8)
= (P2 - P4) (mod 8),
sum of even squares = 0 (mod 8) if M is odd,
= 4 (mod 8) if M is even.
If M is odd, there must be an even number of singly even entries,
if M is even, there must be an odd number of singly even entries,
sum of even entries = 0 (mod 4) if M is odd,
= 2 (mod 4) if M is even.
The conclusion of Argument B is the exact opposite of Argument A: P1 and P3 (mod 4) are inconsistent with P2 and P4 (mod 8).
Therefore, there is no order 25 tetramagic series with 1 odd entry and 24 even entries.
A tetramagic series must then consist of 17 odd and 8 even entries. But 24 rows of 8 even entries covers only 192 of the 312 that need to be covered.
Therefore a 25x25 normal tetramagic square is impossible.
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