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Proof Without Words: Squares Modulo 3Author(s): Roger B. NelsenSource: The College Mathematics Journal, Vol. 44, No. 4 (September 2013), p. 283Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/10.4169/college.math.j.44.4.283 .
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Proof Without Words: Squares Modulo 3
Roger B. Nelsen ([email protected]), Lewis & Clark College, Portland OR97219
n2= 1+ 3+ 5+ · · · + (2n − 1)⇒ n2
≡
{0 (mod 3), n ≡ 0 (mod 3)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 n2, we show vi-sually that n2
≡ 0 (mod 3) when n ≡ 0 (mod 3), and n2≡ 1 (mod 3) when n ≡ ±1
(mod 3).
http://dx.doi.org/10.4169/college.math.j.44.4.283MSC: 00A05, 11-01
VOL. 44, NO. 4, SEPTEMBER 2013 THE COLLEGE MATHEMATICS JOURNAL 283
This content downloaded from 143.88.66.66 on Mon, 9 Sep 2013 20:51:46 PMAll use subject to JSTOR Terms and Conditions
