Exercise 2: Verifying Fermat's Little Theorem for p=11, a=4 — Possible Solution
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GIVEN
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p = 11, a = 4. Set to check: {4, 8, 12, 16, 20, 24, 28, 32, 36, 40}
(that is, {4k} for k = 1 to 10), each reduced mod 11.

STEP 1: COMPUTE {4k mod 11} FOR k = 1 TO 10
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4*1=4   mod 11 = 4
4*2=8   mod 11 = 8
4*3=12  mod 11 = 1
4*4=16  mod 11 = 5
4*5=20  mod 11 = 9
4*6=24  mod 11 = 2
4*7=28  mod 11 = 6
4*8=32  mod 11 = 10
4*9=36  mod 11 = 3
4*10=40 mod 11 = 7

Resulting set: {4, 8, 1, 5, 9, 2, 6, 10, 3, 7}

STEP 2: CONFIRM IT'S A PERMUTATION
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Sorting the resulting set gives {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} -
exactly the original set {1,...,10}, just reordered. Every value
from 1 to 10 appears exactly once, with no repeats and no zero -
confirming this chapter's own permutation claim for this specific
a and p.

STEP 3: CONFIRM THE THEOREM DIRECTLY
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4^10 mod 11 = 1

RESULT
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The permutation property holds exactly as this chapter's proof
requires, and Fermat's Little Theorem's own conclusion (a^(p-1) = 1
mod p) is confirmed directly: 4^10 mod 11 = 1.

WHY THIS WORKS AS AN ANSWER
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Both the intermediate permutation claim (the actual mechanism the
proof depends on) and the theorem's own final conclusion are
verified separately and explicitly, rather than only checking the
end result and assuming the permutation property held along the
way.