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Proof Without Words: Squares Modulo 3 Roger B. Nelsen ([email protected]), Lewis & Clark College, Portland OR 97219 ( 0 (mod 3), n ≡ 0 (mod 3) 2 2 n = 1 + 3 + 5 + · · · + (2n − 1) ⇒ n ≡ 1 (mod 3), n ≡ ±1 (mod 3)

(3k – 1)2 = 1 + 3[(2k – 1)2 – (k – 1)2]

(3k)2 = 3[(2k)2 – k2]

(3k + 1)2 = 1 + 3[(2k + 1)2 – (k + 1)2] Summary. Using the fact that the sum of the first n odd numbers is n 2 , we show visually that n 2 ≡ 0 (mod 3) when n ≡ 0 (mod 3), and n 2 ≡ 1 (mod 3) when n ≡ ±1 (mod 3). http://dx.doi.org/10.4169/college.math.j.44.4.283 MSC: 00A05, 11-01

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