Those Fascinating Numbers 39
120
the smallest tri-perfect number (n is tri-perfect if σ(n) = 3n): only six tri-
perfect numbers are known, namely 120, 672, 523 776, 459 818 240, 1 476 304 896
and 51 001 180 160, and it seems that there are no others (see R.K. Guy [101],
B2);
the only number n which can be joined with the numbers 1, 3 and 8 to form the
set A = {1, 3, 8, n} so that if x, y A, x = y, then xy + 1 is a perfect square;
this result was obtained by Euler; for a thorough analysis of this problem, see
L. Jones [113] or the more recent
paper47
of Dujella [73];
one of the five numbers (the others being 1, 10, 1 540 and 7 140) which are both
triangular and tetrahedral (see the number 10);
the smallest solution of σ2(n) = σ2(n + 10): the only solutions n 108 of this
equation are 120, 942, 5 395, 4 737 595, 6 811 195, 11 151 355, 74 699 995 and
98 600 035.
121
the smallest number n 1 which is
both48
a star number and a perfect square:
a star number is a number of the form 6n(n + 1) + 1; the sequence of num-
bers satisfying this property begins as follows: 121, 11881, 1164241, 114083761,
11179044361, 1095432263641, . . . ;
the only known perfect square of the form 1+p + p2 + p3 + p4, where p is prime:
here with p = 3.
122
the only known number whose square is the sum of a fourth power and a fifth
power: here
1222
=
114
+
35
(see H. Darmon & A. Granville [42] as well as
the number 21 063 928); in fact it is conjectured that the only co-prime integer
solutions x, y, z (non zero) of the equation
xp+yq
=
zr
, with
1
p
+
1
q
+
1
r
1, where
exactly49 one of the numbers p, q, r is equal to 2, are those appearing in the
47A
set of m positive integers {a1, a2, . . . , am} is called a diophantine m-tuple if aiaj + 1 is a
perfect square for all 1 i j m. The first diophantine quadruplet, that is {1, 3, 8, 120},
was found by Fermat. In 1969, Baker & Davenport [11] proved that this quadruplet could not
be extended to a diophantine quintuplet. Let us mention that in 1979, Arkin, Hoggatt & Strauss
[7] proved that each diophantine triplet could be extended to a diophantine quadruplet: indeed,
if {a, b, c} is such a triplet and if ab + 1 =
r2,
ac + 1 =
s2
and bc + 1 =
t2,
where r, s, t are
positive integers, then one easily verifies that d = a + b + c + 2abc + 2rst is such that {a, b, c, d}
is a diophantine quadruplet. In 2004, Dujella [73] proved that no diophantine 6-tuple exists and
that there can only exist a finite number of diophantine 5-tuples, and in fact that any element of a
diophantine 5-tuple must be smaller than
101026
. Let us add that it is easy to prove that there exist
infinitely diophantine quadruplets; indeed, one only needs to prove that there exist infinitely many
diophantine triplets and to use the result of Arkin, Hoggatt & Strauss mentioned above; one then
only needs to verify that the triplets {1,
r2
1,
r2
+ 2r}, where r = 2, 3, 4, . . ., are all diophantine.
48One can establish the recurrence formula Ek = 98Ek−1 Ek−2 + 24, where Ek stands for the
kth
number which is both a star number and a perfect square.
49According to the Beal Conjecture, there are no solutions with min(p, q, r) 3.
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