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The construction of Rijndael's S-box starts with the multiplicative inverses of each number over a finite field. It seems like this is the main source of its non-linearity. I figured that modular multiplicative inverses could be used as well. In this case, determining the value of $S(x)$ starts out with finding $y$ such that $((x + 1) \times (y + 1)) \bmod 257 - 1 = 1$.

Here is the result:

00 80 55 C0 66 2A 92 E0 C7 B3 BA 95 B1 C9 77 F0
78 63 E5 59 30 DD BD 4A 47 58 ED 64 C2 3B C6 F8
93 BC EA 31 83 72 90 2C A2 98 05 6E 27 5E AE A5
14 23 7D AC 60 76 F2 B2 F7 E1 3C 1D 3A E3 65 FC
56 49 E9 DE 94 F5 B4 18 A8 41 17 B9 F6 C8 F3 96
A4 D1 5F CC 7E 02 40 B7 19 13 D0 AF 97 D7 2D 52
34 8A 86 11 1B 3E 04 D6 A3 B0 F4 BB DF F9 2B D9
73 7B 25 70 85 9E 35 0E 10 9D 8B 71 DB 32 54 FE
01 AB CD 24 8E 74 62 EF F1 CA 61 7A 8F DA 84 8C
26 D4 06 20 44 0B 4F 5C 29 FB C1 E4 EE 79 75 CB
AD D2 28 68 50 2F EC E6 48 BF FD 81 33 A0 2E 5B
69 0C 37 09 46 E8 BE 57 E7 4B 0A 6B 21 16 B6 A9
03 9A 1C C5 E2 C3 1E 08 4D 0D 89 9F 53 82 DC EB
5A 51 A1 D8 91 FA 67 5D D3 6F 8D 7C CE 15 43 6C
07 39 C4 3D 9B 12 A7 B8 B5 42 22 CF A6 1A 9C 87
0F 88 36 4E 6A 45 4C 38 1F 6D D5 99 3F AA 7F FF

I understand that this by itself would make for a terrible S-box given its symmetry and fixed points, but perhaps some other math stacked on top of it would make it better?

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  • $\begingroup$ Fundamentally, irreducible polynomials are used for S-Boxes as they cannot be factored. Your example shows some polynomial reductions. Even if you mask over it with something else to remove the fixed points, it will be more complex than using an irreducible polynomial. What was the motivation for this over an irreducible polynomial-based S-Box? $\endgroup$ – b degnan Nov 29 '17 at 4:09
  • $\begingroup$ @bdegnan I figured that modular multiplicative inverses were a good way to start given their non-linearity and that they're the normal arithmetic equivalent of multiplicative inverses in a finite field. $\endgroup$ – Melab Nov 29 '17 at 11:48
  • $\begingroup$ could you please explain what is x and y in the equation ((X+1)*(Y+1)) mod 257 -1 =1 $\endgroup$ – hardyrama May 4 at 6:37

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