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.