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I checked all 4^4 possibilities, but I can’t find any functions where this is true other than the constant functions.
So either my calculations are flawed, or something trickier is happening to b that can't be achieved with multiplication.
Checking all (8!)^4 choices of four independent permutations of ℤ/8 isn’t practical ... and though I *think* the a=1 case must correspond to multiplication by 3, 5 or 7, even 3 * (8!)^3 is too big.
Greg Egan /
npub1qq…j5nhv
2025-03-15 23:03:42
in reply to nevent1q…glv9
Author Public Key
npub1qqm7nu2qf25xd3mw6y6cyp4v2wr7ktf43ymp5wqz4u8jvz76ymtsjj5nhvSeen on
wss://relay.mostr.pubPublished at
2025-03-15 23:03:42Event JSON
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