During the week 13-18 June,2021, contributions came from Giorgos
Kalogeropoulos, Ivan Ianakiev, Emmanuel Vantieghem, Oscar Volpatti.
The problem of puzzle 1041 is mentioned in Richard K. Guy's
"Unsolved Problems in Number Theory", section B13.
On page 104 (of the 3d edition ) we read :
The only solution of s2(n) = s2(n + 1) is n = 6, since
(1) s2(2n) > s2(2n + 1) for n > 7 and
(2) s2(2n) > 5n^2 > ((Pi^2)(2n - 1)^2)/8 > s2(2n - 1)
These inequalities are not evident to me ; I guess they rely on
deeper estimates for sigma2.
Perhaps their proof can be found in :
J.M. De Koninck, On the solutions of Sigma2(n) = Sigma2(n +
l ), Ann. Univ. Sci. Budapest. Sect. Comput., 21 (2002),
(reference taken from Guy's book).
Maybe this reference can be useful to see the 'difficulty'
of the maths involved :
We can prove that
there are no other solutions.
The ratio f(n) =
sigma2(n) / n^2 is bounded for every index n:
A <= f(n) < B
A = 1
B = (pi^2)/6 ~ 1.6449
Sharper bounds can be
found if we distinguish between even and odd indexes:
A1 <= f(2*m-1) < B1
A1 = A = 1
B1 = (pi^2)/8 ~ 1.2337
A2 <= f(2*m) < B2
A2 = 5/4 = 1.25
B2 = B = (pi^2)/6 ~
As A1 < B1 < A2 < B2,
the "odd<even" inequality chain holds for every index m:
B1*(2*m-1)^2 < A2*(2*m)^2 <= sigma2(2*m)
For m > 75.93, the
"even>odd" inequality chain holds too:
A2*(2*m)^2 > B1*(2*m+1)^2 > sigma2(2*m+1)
So we can solve
sigma2(n) = sigma2(n+1) only if n = 2*m and n+1 = 2*m+1 and m <=
provides only the known solution for m = 3.
Bound A = A1 = 1
The largest divisor of
an integer n is n itself, so sigma2(n) >= n^2
Bound A2 = 5/4
The largest divisors
of an even integer n are n itself and n/2, so sigma2(n) >= n^2 +
Bound B = B2 =
If we know the
factorization of n as a product of prime powers p_k^e_k,
then we can compute
sigma2(n) as a product of geometric sums Q_k:
Q_k = 1 + p_k^2 + ...
and the ratio f(n) =
sigma2(n) / n^2 as a product of geometric sums R_k = Q_k/(p_k^(2*e_k)):
R_k = 1 + p_k^(-2) +
... + p_k^(-2*e_k)
An upper bound for R_k,
not depending on e_k, is given by replacing the finite sum with
a converging geometric series:
R_k < 1 + p_k^(-2) +
p_k^(-4) + ... = (p_k^2)/(p_k^2 - 1)
A further upper bound
on f(n) is given by replacing the finite product, extended to
the prime factors p_k of n, with a converging infinite product,
extended to all primes:
(4/3)*(9/8)*(25/24)*(49/48)*(121/120)*... = zeta(2) =
(pi^2)/6 ~ 1.6449
Euler proved the
product formula for Riemann zeta function and the closed-form
evaluation of zeta(2).
Bound B1 = (p1^2)/8
If we consider only
odd integers, we must extend the infinite product to odd primes
only, reducing the upper bound by the factor 4/3.