Proof Without Words: Perfect Numbers Modulo 7 17 - Google Sites

Proof Without Words: Perfect Numbers Modulo 7 Roger B. Nelsen ([email protected]), Lewis & Clark College, Portland, OR Theorem ([1]). Every even perfect number N p = 2 p−1 (2 p − 1) for prime p = 3 is congruent to 1 or 6 modulo 7. In particular, p ≡ 1 mod 3 =⇒ N p ≡ 1 mod 7

and

p ≡ 2 mod 3 =⇒ N p ≡ 6 mod 7.

Proof. N p = 2 p−1 (2 p − 1) = T2 p −1 where Tn = 1 + 2 + · · · n = n(n + 1)/2 is the nth triangular number:

2 p–1–1 p

2 –1

2

2

p–1

p–1

p

2 –1

p = 3k + 1 =⇒ 2 p − 1 = 2 · 8k − 1 ≡ 1 mod 7 =⇒ N3k+1 = T7n+1 , p = 3k + 2 =⇒ 2 p − 1 = 4 · 8k − 1 ≡ 3 mod 7 =⇒ N3k+2 = T7n+3 :

T7n+1 = 35Tn + 14Tn−1 + 1,

T7n+3 = 49Tn + 6,

so N3k+1 ≡ 1 mod 7.

so N3k+2 ≡ 6 mod 7.

Summary. We partition triangular numbers to show wordlessly that every even perfect number except 28 is congruent to 1 or 6 modulo 7. Reference 1. C. R. Wall, Even perfect numbers mod 7, Fibonacci Quart. 22 (1984) 274–275. http://dx.doi.org/10.4169/college.math.j.48.1.17 MSC: 11A07

VOL. 48, NO. 1, JANUARY 2017 THE COLLEGE MATHEMATICS JOURNAL

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