Solution: Set $ x = y = 1 $: $ k(2) = k(1) + k(1) - 2k(1 \cdot 1) = 1 + 1 - 2(1) = 0 $. Verify consistency: Try $ x = 2, y = 1 $: $ k(3) = k(2) + k(1) - 2k(2) = 0 + 1 - 0 = 1 $. Try $ x = y = 2 $: $ k(4) = k(2) + k(2) - 2k(4) $ â†’ $ k(4) + 2k(4) = 0 + 0 $ â†’ $ 3k(4) = 0 $ â†’ $ k(4) = 0 $. Assume $ k(x) = 0 $ for all $ x $, but $ k(1) = 1

Sun, 03 May 2026 05:20:27 +0000

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