On the Solutions of σ(n) = σ(n + k)

Andreas Weingartner Department of Mathematics Southern Utah University Cedar City, Utah 84720 USA

                      weingartner@suu.edu

Abstract Given a fixed even integer k, we show that Schinzel’s hypothesis H implies that σ(n) = σ(n + k) infinitely often. We also discuss the case of odd k and the more general equation σ_{α}(n) = σ_{α}(n + k).

1 Introduction

Let σ(n) be the sum of the positive divisors of n. Sierpiński [9, p. 166] asked whether

σ(n) = σ(n + 1)  (1)

infinitely often. The sequence of solutions (OEIS \underline{A002961}) begins with

14, 206, 957, 1334, 1364, 1634, 2685, 2974, 4364, 14841, ...

Although Sierpiński’s question has not been answered yet, computational results [5, 6, 7] seem to suggest that the sequence may have infinitely many terms. Guy and Shanks [3] constructed a large solution of (1) based on a pattern observed in smaller ones, but they point out that their construction is unlikely to yield an infinite number of solutions.

More generally, Mientka and Vogt [7] asked for which values of k ≥ 1,

σ(n) = σ(n + k)  (2)

has infinitely many solutions. Hunsucker, Nebb and Stearns [6] found at least two solutions of (2) for each k ≤ 5000.

Erdős [2] made the much stronger conjecture that for every integer k ≥ 1 there is an n such that

σ(n) = σ(n + 1) = ⋯ = σ(n + k),

which would clearly imply that (2) has infinitely many solutions for each k. However, a computer search showed that there is no solution to σ(n) = σ(n+1) = σ(n+2) for n ≤ 10^{10}.

Let f_{1}(x), f_{2}(x), …, f_{n}(x) be irreducible polynomials with integer coefficients and positive leading coefficients. Schinzel’s hypothesis H [8] says that, if there is no fixed integer greater than one which divides

f(x) = f_{1}(x)f_{2}(x) ⋯ f_{n}(x)

for all integers x, then f_{1}(x), …, f_{n}(x) are simultaneously prime for infinitely many values of x.

In Section 2 we show that if k is even, the existence of an infinite number of solutions of (2) follows from a particular case of Schinzel’s hypothesis H. For k = 1 we show that if odd multiperfect numbers do not exist, then there are only finitely many solutions n to (1) with Ω(n) + Ω(n + 1) bounded by an absolute constant, where Ω(n) denotes the number of prime divisors of n, counted with multiplicity.

In Section 3 we take a look at the corresponding questions for σ_{α}(n), the sum of the α-th powers of the positive divisors of n.

2 The case α = 1.

The first ten solutions to (2) for k = 2 are (OEIS \underline{A007373})

33, 54, 284, 366, 834, 848, 918, 1240, 1504, 2910, ....

The following result shows that this sequence, and all other sequences corresponding to even values of k (\underline{A015863} (k = 4), \underline{A015866} (k = 6), \underline{A015876} (k = 8), \underline{A015880} (k = 10), \underline{A015882} (k = 12), \underline{A181647} (k = 20)), are infinite sequences, provided Schinzel’s hypothesis H holds.

Proposition 1. Let k be a fixed even integer. Assuming Schinzel’s hypothesis H, the equation σ(n) = σ(n + k) has infinitely many solutions.

Proof. We begin with the case k = 2. Define

p = 2x + 1, q = 3x + 8, r = 2x + 5, s = 3x + 2,  (3)
n = pq, m = rs,

and

f(x) = pqrs = (2x + 1)(3x + 8)(2x + 5)(3x + 2).

The function f(x) does not have a fixed divisor > 1, since, for example, f(0) = 2^{4} · 5 and f(3) = 7 · 11^{2} · 17 have no common factor. Schinzel’s hypothesis H says that there are infinitely many values of x such that each one of the four numbers p, q, r, and s is prime. For these values of x we have σ(n) = σ(m), since

σ(n) = (p + 1)(q + 1) = 6(x + 1)(x + 3) = (r + 1)(s + 1) = σ(m).

Furthermore, m = n + 2 for all x. This completes the case k = 2.

If k = 2l with l > 1, let m and n be as above, and write

n′ = ln, m′ = lm.

We have m′ = n′ + 2l and

σ(m′) = σ(l)σ(m) = σ(l)σ(n) = σ(n′),

if gcd(l, mn) = 1, which is the case as soon as p = 2x + 1 > l.

Without Schinzel’s hypothesis H, we have the following unconditional result.

Proposition 2. The equation σ(n) = σ(n + k) has at least one solution for every even k with 2 ≤ k ≤ 10^{10^{7}}.

Proof. Using a computer, we generate a sequence 33 = x_{1} < x_{2} < ⋯ < x_{N} = 146344933173, with N = 913685 terms, such that p_{i} = 2x_{i} + 1, q_{i} = 3x_{i} + 8, r_{i} = 2x_{i} + 5, and s_{i} = 3x_{i} + 2 are distinct primes for 1 ≤ i ≤ N. It is easy to see that x_{i} ≡ 3 (mod 30) if p_{i}, q_{i}, r_{i} and s_{i} are not divisible by 2, 3, or 5. To ensure that we obtain 4N distinct primes in all, we further restrict the search to x_{i} ≡ 33 (mod 60). As in the proof of Proposition 1, we have σ(p_{i}q_{i}) = σ(r_{i}s_{i}) = σ(p_{i}q_{i} + 2), and therefore σ(p_{i}q_{i}l) = σ(p_{i}q_{i}l + 2l) if gcd(l, p_{i}q_{i}r_{i}s_{i}) = 1, where k = 2l. If this fails for all i ≤ N, then l must be divisible by at least one of the four primes p_{i}, q_{i}, r_{i}, or s_{i}, for each i ≤ N. Thus k > l ≥ ∏_{i=1}^{N}(2x_{i} + 1) > 10^{10^{7}}.

For odd values of k, the situation is quite different. Assume that there is a sequence n_{j} of solutions to (2) of the form

n_{j} = a ∏_{i=1}^{N} p_{i,j}^{α_{i,j}}, n_{j} + k = b ∏_{i=1}^{M} q_{i,j}^{β_{i,j}},  (4)

where a, b, N, M are fixed and the primes p_{i,j}, q_{i,j} satisfy lim_{j→∞} p_{i,j} = ∞ for 1 ≤ i ≤ N and lim_{j→∞} q_{i,j} = ∞ for 1 ≤ i ≤ M. Then, as n_{j} grows,

\frac{σ(n_{j})}{n_{j}} = \frac{σ(a)}{a} ∏_{i=1}^{N} \frac{p_{i,j}^{α_{i,j}+1} − 1}{(p_{i,j} − 1)p_{i,j}^{α_{i,j}}} = \frac{σ(a)}{a}(1 + o(1)),

\frac{σ(n_{j} + k)}{n_{j} + k} = \frac{σ(b)}{b} ∏_{i=1}^{M} \frac{q_{i,j}^{β_{i,j}+1} − 1}{(q_{i,j} − 1)q_{i,j}^{β_{i,j}}} = \frac{σ(b)}{b}(1 + o(1)).

Since σ(n_{j}) = σ(n_{j} + k), we also have

\frac{σ(n_{j})}{n_{j}} = \frac{σ(n_{j} + k)}{n_{j} + k} \frac{n_{j} + k}{n_{j}} = \frac{σ(n_{j} + k)}{n_{j} + k}(1 + o(1)).

It follows that

\frac{σ(a)}{a} = \frac{σ(b)}{b},  (5)

where one of a and b is even, and the other one is odd, since k is odd. When k = 1 we also have gcd(a, b) = 1, which means that the fractions in (5) must reduce to integers ≥ 2. Thus a and b are both multiperfect numbers, with one of them being odd. Since no odd multiperfect numbers are known, we can not use this approach to show that there are infinitely many solutions to (1). We summarize this in the following statement.

Proposition 3. Let a, b, N, M be fixed positive integers, such that neither a nor b is an odd multiperfect number, and let n_{j} be a sequence of the form (4) with k = 1, where the primes p_{i,j}, q_{i,j} satisfy lim_{j→∞} p_{i,j} = ∞ for 1 ≤ i ≤ N and lim_{j→∞} q_{i,j} = ∞ for 1 ≤ i ≤ M. Then the sequence n_{j} contains at most a finite number of solutions to (1).

The following result is a consequence of Proposition 3.

Proposition 4. Let C ≥ 1 be fixed. If there are no odd multiperfect numbers, then there are only finitely many solutions to σ(n) = σ(n + 1) with Ω(n) + Ω(n + 1) ≤ C.

Proof. Assume that n_{j} is an infinite increasing sequence with σ(n_{j}) = σ(n_{j} + 1) and Ω(n_{j}) + Ω(n_{j} + 1) ≤ C. We show that n_{j} has an infinite subsequence of the form (4) with k = 1. By restricting n_{j} to a subsequence if necessary, we may assume Ω(n_{j}) = μ and Ω(n_{j} +1) = ν, for some constants μ, ν ∈ ℕ. Write n_{j} = p_{1,j} ⋯ p_{μ,j} and n_{j} +1 = q_{1,j} ⋯ q_{ν,j}. For each 1 ≤ i ≤ μ, either lim inf_{j→∞} p_{i,j} = p_{i} for some fixed prime p_{i}, or lim_{j→∞} p_{i,j} = ∞. In the first case we restrict the sequence n_{j} to an infinite subsequence such that p_{i,j} = p_{i} for all j ≥ 1. Let a denote the product of all the fixed primes p_{i}. After applying the same process to the prime divisors of n_{j} + 1, we arrive at an infinite sequence of the form (4) with k = 1. Thus the result follows from Proposition 3.

For odd values of k ≥ 3, it may still be possible to construct an infinite sequence of solutions of the form (4), because there are solutions to (5) with a even and b odd, such as

a = 3472 = 2^{4} · 7 · 31, b = 544635 = 3^{2} · 5 · 7^{2} · 13 · 19.

In this case gcd(a, b) = 7, so it may be possible to use this as a starting point to show that σ(n) = σ(n + 7) has infinitely many solutions, assuming Schinzel’s hypothesis H, although we were not able to do so.

While we can’t prove or disprove that there is at least one solution of (2) for every odd value of k, in some cases a solution is easy to find. For example, n = 14k is a solution to (2) if gcd(k, 14 · 15) = 1, since n = 14 is a solution to (1). The same idea can be applied to other solutions of (1).

3 The case α ≥ 2.

The case α = 2 was treated by De Koninck [1] with the following result.

Proposition 5 (De Koninck). Let k be a fixed positive integer.

(i) If k is odd, then σ_{2}(n) = σ_{2}(n + k) has at most a finite number of solutions.

(ii) If k is even, then Schinzel’s hypothesis H implies that σ_{2}(n) = σ_{2}(n+k) has an infinite number of solutions.

The only solution to σ_{2}(n) = σ_{2}(n + 1) is n = 6. According to [1], σ_{2}(n) = σ_{2}(n + k) has no solution when k = 3, 9, 15, 27, 33, 35, 39, 45, 51, 57, 69, 75, 81, 87, 93 or 99. Assuming Schinzel’s hypothesis H, a sequence of solutions to σ_{2}(n) = σ_{2}(n + 2) given in [1] can be written as

n = x^{3} − 1 = (x − 1)(x^{2} + x + 1), n + 2 = x^{3} + 1 = (x + 1)(x^{2} − x + 1).

It is easy to verify that σ_{2}(n) = σ_{2}(n+2) as long as the four quantities x−1, x^{2} +x+1, x+1 and x^{2} − x + 1 are all prime.

We can extend the first half of Proposition 5 to α ≥ 3.

Proposition 6. For every α ≥ 2 there is a constant c_{α} > 0 such that if σ_{α}(n) = σ_{α}(n + k) for an odd value of k, then k > c_{α}n.

Proof. We first show that

1 + \frac{1}{2^{α}} > ∑_{j≥0} \frac{1}{(2j + 1)^{α}}  (α ≥ 2).  (6)

For α = 2, (6) simplifies to \frac{5}{4} > (1 − 1/2^{2})ζ(2) = \frac{π^{2}}{8} = 1.2337... If we subtract 1 from both sides of (6) and then multiply by 2^{α}, we see that (6) is equivalent to

1 > ∑_{j≥1} \frac{1}{(j + 1/2)^{α}} (α ≥ 2).

Note that the right-hand side is decreasing in α. Since the inequality is valid for α = 2, it is valid for all α ≥ 2.

Now let m = n+k. Since k is odd, one of m, n is even and the other one is odd. Without loss of generality we assume 2|n and 2 |̸m, allowing k < 0 if necessary. Then

\frac{σ_{α}(n)}{n^{α}} ≥ \frac{σ_{α}(2)}{2^{α}} = 1 + \frac{1}{2^{α}} ,

and

\frac{σ_{α}(m)}{m^{α}} < ∑_{j≥0} \frac{1}{(2j + 1)^{α}} .

If σ_{α}(n) = σ_{α}(m),

(1 + \frac{1}{2^{α}} ) (\frac{n}{n + k} )^{α} ≤ \frac{σ_{α}(n)}{n^{α}} \frac{n^{α}}{m^{α}} = \frac{σ_{α}(m)}{m^{α}} < ∑_{j≥0} \frac{1}{(2j + 1)^{α}} ,

or

(\frac{1}{1 + k/n})^{α} < \frac{1 − \frac{1}{2^{α}}}{1 + \frac{1}{2^{α}}} ζ(α) =: b_{α},

say. From (6) we have b_{α} < 1 for α ≥ 2, hence

\frac{k}{n} > b_{α}^{−1/α} − 1 =: c_{α} > 0.

The remaining case is where α ≥ 3 and k is even. Guy [4, Problem B13] writes, “Erdős thinks that σ_{3}(n) = σ_{3}(n + 2) has no solution at all”. A computer search showed that there is in fact no solution to σ_{3}(n) = σ_{3}(n + 2) for n ≤ 10^{10}. Although we are not able to prove that there are no solutions in this case, we have the following partial result. We omit the proof since the ideas are similar to those of Proposition 6.

Proposition 7. If n is a solution of σ_{3}(n) = σ_{3}(n + 2), then neither n nor n + 2 is divisible by 2, 3, 5 or 7.

For α ≥ 3, solutions to σ_{α}(n) = σ_{α}(n + k) are quite rare even if k is allowed to vary. There are only three primitive solutions to σ_{3}(n) = σ_{3}(n + k) with n ≤ 5 · 10^{6} and k ≥ 1, namely

(n, k) = (184926, 9389), (291741, 3560), (1880574, 6346),

while the other fourteen solutions in the same range are obtained from these by multiplying both n and k by some l with gcd(l, nk) = 1.

There are no solutions to σ_{α}(n) = σ_{α}(n + k) for α ≥ 4, n ≤ 10^{6} and k ≥ 1.

References

[1] J.-M. De Koninck, On the solutions of σ_{2}(n) = σ_{2}(n + l), Ann. Univ. Sci. Budapest. Sect. Comput. 21 (2002), 127–133.

[2] P. Erdős, Some remarks on Euler’s ϕ function and some related problems, Bull. Amer. Math. Soc. 51 (1945), 540–544.

[3] R. K. Guy and D. Shanks, A constructed solution of σ(n) = σ(n + 1), Fibonacci Quart. 12 (1974), 299.

[4] R. K. Guy, Unsolved Problems in Number Theory, Springer, 2004.

[5] P. Haukkanen, Some computational results concerning the divisor functions d(n) and σ(n), Math. Student 62 (1993), 166–168.

[6] J. L. Hunsucker, J. Nebb, and R. E. Stearns, Computational results concerning some equations involving σ(n), Math. Student 41 (1973), 285–289.

[7] W. E. Mientka and R. L. Vogt, Computational results relating to problems concerning σ(n), Mat. Vesnik 7 (1970), 35–36.

[8] A. Schinzel and W. Sierpiński, Sur certaines hypothèses concernant les nombres premiers, Acta Arith. 4 (1958), 185–208.

[9] W. Sierpiński, Elementary Theory of Numbers, Polska Akademia Nauk, 1964.

2010 Mathematics Subject Classification: Primary 11A25. Keywords: Sum-of-divisors function, Schinzel’s hypothesis H.

(Concerned with sequences \underline{A002961}, \underline{A007373}, \underline{A015863}, \underline{A015866}, \underline{A015876}, \underline{A015880}, \underline{A015882}, and \underline{A181647}.)

Received February 26 2011; revised version received May 2 2011. Published in Journal of Integer Sequences, May 3 2011.

Return to Journal of Integer Sequences home page.