334 Jean-Marie De Koninck
731 601 000
the eighth number n such that φ(n) + σ(n) = 4n (see the number 23 760).
740 461 601
the smallest number n which allows the sum
i≤n
1
i
to exceed 21 (see the number
83).
745 988 807
the smallest prime factor of the Mersenne number 2109 1, whose complete
factorization is given by
2109
1 = 745 988 807 · 870 035 986 098 720 987 332 873.
746 444 160 (= 27 · 34 · 5 · 7 · 112 · 17)
the smallest solution of
σ(n)
n
=
19
4
.
776 151 559 (=
9193)
the smallest 3-powerful number which can be written as the sum of two co-
prime 3-powerful numbers: here
9193
=
2713
+
23
·
35
·
733;
A. Nitaj [153]
proved a conjecture of Erd˝ os going back to 1975 which claims that there exist
infinitely
many199
triplets of 3-powerful numbers a, b, c, with (a, b) = 1, such
that a + b = c; this by the way reveals a second 3-powerful number which can
be written as the sum of two 3-powerful numbers, namely the number
11205183603973252067=373
·
1973
·
3073 =27
·
34
·
53
·
73
·
22873
+
173
·
1062193;
however, there exist other 3-powerful numbers with such a representation, such
as the number 3518958160000, for which
3518958160000 =
27
·
54
·
3533
=
34
·
293
·
893
+
73
·
113
·
1673.
199Indeed,
this comes from the fact that equation (∗)
x3
+
y3
=
6z3
has infinitely many solutions.
To prove this, first observe that one can generate infinitely many solutions (xk, yk, zk) of (∗) by
considering the sequence defined by x0 = 37, y0 = 17, z0 = 21 and for each k 0, by
xk+1 = xk(xk
3
+ 2yk),
3
yk+1 = −yk(2xk
3
+ yk),
3
zk+1 = zk(xk
3

yk).3
In order to show that the numbers a = x3
k
, b = y3
k
and c = 6z3
k
are adequate, that is 3-powerful for
each k 1, it remains to prove that zk is divisible by 6 for each integer k 1. But this is trivial.
Indeed, on the one hand, since xk and yk are odd for each k 0, it follows that zk is even for each
k 1. On the other hand, since 3|z0, then 3|z1, implying that all the other zi’s are also multiples
of 3.
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