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

From S2,

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.

From S1,

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)

thus

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

Argument A

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 4)
= (P1 - sum of odd cubes) (mod 4)
= (P1 - P3) (mod 4),

therefore

sum of even entries = 2 (mod 4) if M is odd,
= 0 (mod 4) if M is even.

Argument B

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 8)
= (P2 - sum of odd 4th powers) (mod 8)
= (P2 - P4) (mod 8),

hence

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,

therefore

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.