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.
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