14 Jean-Marie De Koninck
the smallest number n 1 such that n2 +3 is a powerful number: here 372 +3 =
22 · 73; the sequence of numbers satisfying this property begins as follows: 37,
79 196, 177 833, . . . ; Florian Luca proved that this sequence is infinite16, and in
fact his argument reveals in particular two additional solutions of n2 + 3 = m,
where m is powerful, namely
96679390102
+ 3 = 93469044701079780103
=
73
·
612
·
1092
·
785112,
25248079505075105232
+ 3 = 6374655186945935706615682630955733532
=
22
·
77
·
192
·
732152605848612;
Patrick Letendre observes that more generally, one can prove that if there
exists a powerful number of the form n2 + k which is not a perfect square, then
there are infinitely many of them17;
the largest solution y of the diophantine equation
x2
+28 =
y3
(see the number
225);
the smallest number 1 which is equal to the sum of the squares of the
factorials of its digits (in base 12): here 73 = [3, 3, 1]12 =
3!2
+
3!2
+
1!2
(see
the numbers 145 and 40 465); the set of numbers satisfying this property is of
course finite and its four smallest elements are 1, 37, 613 and 519 018;
the largest known prime number which is also a convenient number; an odd
number n 1 is said to be convenient if it has only one representation of
the form n = x2 + my2, with x, y positive and (x, my) = 1; the only known
convenient prime numbers are 2, 3, 5, 7, 13 and 37 (see also the number 1 848).
38
the largest even number which cannot be written as the sum of two odd com-
posite
numbers18.
16Here
is Luca’s argument. First consider the Fermat-Pell equation
x2

7y2
= 1. It has infinitely
many solutions (xn, yn), n = 1, 2, . . ., the smallest of which is (x1, y1) = (8, 3). For each of these
solutions (xn, yn), consider the numbers x = 37xn + 98yn and y = 14xn + 37yn. One easily verifies
that x2 + 3 = 7y2. To complete the proof, one needs to have that y is a multiple of 7. But y 2yn
(mod 7), so that if yn is a multiple of 7, the same will be true for y. Since it is the case for
n = 7, 14, 21, . . ., the result follows.
17Let
d be a positive integer which is not a perfect square and consider the quantities n =
x2 −dy2
and m =
a2

db2.
One easily verifies that (∗) m · n = (ax
dby)2
d(ay +
bx)2.
This is why,
knowing numbers x0 and y0 such that x0
2
dy0
2
= 1 (it is known that there are infinitely many
of these), and starting with the known solution (t, z) of equation
t2
+ k = z, with z powerful (=
perfect square), then let
u2
be the largest square divisor of z satisfying z =
u2
· v, where (u, v) = 1
and v 1. This brings us to equation
t2

vu2
= −k, of which we already know by hypothesis a
solution (t0, u0), which in turn allows us to generate infinitely many solutions using (∗).
18The
proof of this result is very simple. Indeed, let n = 2m be an even number. This number m
is necessarily of one of the three following forms: m = 3k, m = 3k+1 or m = 3k+2. In the first case,
n = 2m = 9 + 3 ·
2m−3
3
, the sum of two odd composite numbers. In the second case, we can write
n = 2m = 2(3k+1) = 35+3(2k−11). In the third case, we have n = 2m = 2(3k+2) = 25+3(2k−7).
We have thus settled the case of all even numbers which are multiples of 3 and 9 + 6 = 15, that
of each number 2m with m = 3k + 1 and 2m 35 + 3 · 3 = 44 and that of each number 2m with
m = 3k + 2 and 2m 25 + 3 · 3 = 34. Therefore, since 40 = 15 + 25 and since we can easily verify
that 38 is not the sum of two odd composite numbers, the result follows.
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