Problems & Puzzles: Puzzles

 Puzzle 1041. σ2(x)=σ2(x+1) Fred Schneider sent the following nice puzzle.Let's define σ2(x) as "the sum of the square of the divisors of x". Now observe that: σ2(6)=1^2+2^2+3^2+6^2=50, and σ2(7)=1^2+7^2=50 Q. Find another x such that σ2(x)=σ2(x+1) or prove that there are no other solutions.

During the week 13-18 June,2021, contributions came from Giorgos Kalogeropoulos, Ivan Ianakiev, Emmanuel Vantieghem, Oscar Volpatti.

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Giorgios wrote:

It is proved that n=6 is the only solution.
look at the third page of the pdf I am sending you. It is also stated in R.Guys book B13 (see the last paragraph)

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Ivan wrote:

According to J.-M. De Koninck (see link) there is no solution other than 6.

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Emmanuel wrote:

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), 127-133.
(reference taken from Guy's book).

...

Maybe this reference can be useful to see the 'difficulty' of the maths involved :

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Oscar wrote:

We can prove that there are no other solutions.

Step 1.
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

Step 2.
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 ~ 1.6449

Step 3.
As A1 < B1 < A2 < B2, the "odd<even" inequality chain holds for every index m:
sigma2(2*m-1) < B1*(2*m-1)^2 < A2*(2*m)^2 <= sigma2(2*m)
For m > 75.93, the "even>odd" inequality chain holds too:
sigma2(2*m) >= 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 <= 75.
Exaustive search 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 + (n/2)^2

Bound B = B2 = (pi^2)/6
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 + ... + p_k^(2*e_k)
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:
f(n) < (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.

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