Theorem. Every even perfect number Np = 2pâ1(2p â 1) with p ⥠3 prime is the sum of the first n odd cubes for n =
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MATHEMATICS MAGAZINE
Proof Without Words: Perfect Numbers and Sums of Odd Cubes R O G E R B. N E L S E N
Lewis & Clark College Portland, OR
[email protected]
Theorem. Every even perfect number N p = 2 p−1 (2 p − 1) with p ≥ 3 prime is the sum of the first n odd cubes for n = 2( p−1)/2 , i.e., N p = 13 + 33 + · · · + (2n − 1)3 [1]. Proof: (e.g., for p = 5, N5 = 16 · 31 and n = 4): Let Tk = 1 + 2 + · · · + k = k(k + 1)/2 denote the k th triangular number. Then 1. N p = 2 p−1 (2 p − 1) = T2n2 −1 .
2 p–1 – 1
2p – 1
2 p–1
2 p–1
2p – 1 = 2n2 – 1
2. T2n2 −1 = 1 · 12 + 3 · 32 + · · · + (2n − 1) · (2n − 1)2 .
c Mathematical Association of America Math. Mag. 89 (2016) 14–15. doi:10.4169/math.mag.89.1.14. MSC: Primary 11A25, Secondary 05A15
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The theorem actually holds for p ≥ 3 odd, whether or not p or 2 − 1 is prime. p
REFERENCE 1. S. Kahan, Perfectly odd cubes, Math. Mag. 71 (1998) 131. Summary. cubes.
We show wordlessly that every even perfect number greater than six is a sum of consecutive odd
ROGER NELSEN (MR Author ID: 237909) is professor emeritus at Lewis & Clark College, where he taught mathematics and statistics for 40 years.
Artist Spotlight Bjarne Jespersen Flexus, Bjarne Jespersen; Mahogany, 5 cm in diameter, 1971; private collection. Four three-sided hosohedral edge frames weave together. See interview on page 55.
