Those Fascinating Numbers 89
470
the fifth number α such that σ(n) =

for all numbers n; the sequence of
numbers α satisfying this property begins as follows: 1, 4, 6, 11, 470, 475, 477,
480, 482, 483, 484, 485, 486, 487, 488, 489, 490, 491, 492, 493, 494, 495, 496,
497, 498, 499, 500, 501, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 512,
513, 514, 515, 516, 517, 518, 519, 520, 522, 525, 527, 532, . . . 92
479
the smallest prime factor of the Mersenne number
2239
1, whose complete
factorization is given by
2239
1 = 479 · 1913 · 5737 · 176383 · 134000609
·7110008717824458123105014279253754096863768062879.
481
the smallest number whose square can be written as the sum of three fourth
powers:
4812
=
124+154+204;
the ten smallest numbers satisfying this property
are 481, 1 924, 4 329, 7 696, 12 025, 17 316, 23 569, 24 961, 28 721 and 30 784;
possibly the largest number n such that n(n + 1) . . . (n + 5) has exactly the
same prime factors as m(m + 1) . . . (m + 5) for a certain number m n: here
m = 480 and the prime factors common to these two quantities are 2, 3, 5, 7,
11, 13, 23, 37, 97 and 241, since
480 · 481 · . . . · 485 =
28
·
32
·
52
· 7 ·
112
· 13 · 23 · 37 · 97 · 241,
481 · 482 · . . . · 486 =
24
·
36
· 5 · 7 ·
112
· 13 · 23 · 37 · 97 · 241;
see the number 340.
485
the fifth number n such that n2 1 is powerful: here 4852 1 = 23 · 35 · 112; the
sequence of numbers satisfying this property is infinite93 and begins as follows:
3, 17, 26, 99, 485, 577, 1 351, 3 363, 19 601, 24 335, 70 226, 114 243, 470 449,
665 857, 930 249, 2 862 251, 3 650 401, 3 880 899, . . .
487
the second prime number p such that
10p−1
1 (mod
p2):
the only prime
numbers p
232
satisfying this congruence are 3, 487 and 56 598 313 (see
Ribenboim [169], p. 347).
92It
is easy to prove that σ(n) is a power of 2 if and only if n is the product of Mersenne primes.
93This follows from the fact that to each solution (x, y) of the Fermat-Pell equation x2 2y2 = 1,
one can associate a number n such that
n2
1 is powerful. Indeed, choosing n = x, we then have
n2 1 = x2 1 = 2y2, and since y is necessarily even, it follows that 2y2 is powerful.
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