100 Jean-Marie De Koninck

671

• the fifth self contained number (see the number 293);

• the value of the sum of the elements of a diagonal, of a line or of a column in

a 11 × 11 magic square (see the number 15).

672 (=

25

· 3 · 21)

• the second tri-perfect number (see the number 120).

674

• the smallest number which can be written as the sum of four fourth powers as

well as five fourth powers: 674 =

34

+

34

+

44

+

44

=

14

+

24

+

24

+

24

+

54.

675

• the third powerful number n such that n + 1 is also powerful: here 675 = 33 · 52

and 676 = 22 · 132 (see the number 288).

679

• the smallest number of persistence 5: we say that a number is of persistence k

if the number of iterations required in the process of multiplying the digits to

finally arrive at only one digit is k:

679 −→ 378 −→ 168 −→ 48 −→ 32 −→ 6;

if nk stands for the smallest number of persistence k, then n1 = 11, n2 = 25,

n3 = 39, n4 = 77, n5 = 679, n6 = 6 788, n7 = 68 889, n8 = 2 677 889, n9 =

26 888 999 and n10 = 3 778 888 999.

682

• the smallest number n such that

n2

+1 is a

powerful104

number: here

6822

+1 =

53 ·612;

the sequence of numbers satisfying this property begins as follows: 682,

1 268 860 318, 2 360 712 083 917 682, 4 392 100 110 703 410 665 318,

8 171 493 471 761 113 423 918 890 682, . . . , and this sequence is

infinite105.

104It

is easy to see that such a number must be even, since otherwise 2

n2

+ 1 in which case

n2

+ 1

is not powerful.

105Indeed,

since each powerful number is of the form

x2y3

for certain numbers x and y (with

µ2(y)

= 1), we only need to show that equation

n2

+ 1 =

x2y3

has infinitely many integer solutions

n, x, y. But we already know the solution (n, x, y) = (682, 61, 5), meaning that the Fermat-Pell

equation

n2

−

125x2

= −1 has a solution (namely (n, x) = (682, 61) = (n1, x1), say). Therefore, it

follows that equation

n2

−

125x2

= −1 has infinitely many solutions (n2k+1, x2k+1), k = 1, 2, . . .,

given implicitly by

n2k+1 + x2k+1

√

125 = (n1 + x1

√

125)2k+1.