62 Jean-Marie De Koninck

239

• one of the two numbers (the other one is 23) which cannot be written as the

sum of eight cubes (in non negative integers) or less: 239 = 2 · 43 +4 · 33 +3 · 13;

• the largest number n such that P (n2 + 1) 17, where P (m) stands for the

largest prime factor of m: Maurice Mignotte proved that P (n2 + 1) ≥ 17 if

n ≥ 240, while 2392 + 1 = 2 · 134 (R.K. Guy [101], A1); see the number 137;

• the only solution n 1 of the diophantine equation n2 + 1 = 2m4: in 1942,

W. Ljunggren proved that the only two positive integer solutions of n2+1 = 2m4

are (n, m) = (1, 1) and (239,13) (R.K. Guy [101], D6);

240

• the 12th highly composite number (see the number 180);

• the smallest number m for which equation σ(x) = m has exactly seven solutions,

namely 114, 135, 158, 177, 203, 209 and 239;

• the fifth number n 1 such that φ(σ(n)) = n (see the number 128).

241

• the smallest number n such that τ (n) ≤ τ (n + 1) ≤ . . . ≤ τ (n + 4) (as well as

the smallest such that τ (n) ≤ τ (n + 1) ≤ . . . ≤ τ (n + 5)): here 2 6 ≤ 6 ≤ 6 ≤

6 8; if we denote by nk the smallest number n such that τ (n) ≤ τ (n + 1) ≤

. . . ≤ τ (n + k − 1), then we have the following table:

k 2 3 4 5 6 7 8

nk 9 13 13 241 241 12 853 12 853

k 9 10 11 12

nk 234 613 376 741 78 312 721 125 938 261

242

• the smallest number n such that n, n + 1, n + 2 and n + 3 are all divisible by

a square 1: here 242 = 2 · 112, 243 = 35, 244 = 22 · 61 and 245 = 5 · 72; if

we denote by nk the smallest number n such72 that µ(n) = µ(n + 1) = . . . =

µ(n + k − 1) = 0, then n2 = 8, n3 = 48, n4 = 242, n5 = 844, n6 = 22 020,

n7 = 217 070, n8 = 1 092 747, n9 = 8 870 024, n10 = n11 = 221 167 422 and

n12 = 47 255 689 915; if we denote by mk, the smallest number m such that

72It

follows from the Chinese Remainder Theorem that nk exists for each k ≥ 2 and moreover

that nk

k

i=1

pi

2

; similarly, mk, exists for each k ≥ 2 and ≥ 2, and we also have that

mk,

k

i=1

pi .