Do more than three integers $n$ have divisor sums which are odd squares?

In a search for integers $n$ whose divisors sum to a square (including $n$ itself among the divisors), it can be shown that 3 is the only odd prime whose divisor sum is square, and that no square of an odd prime has a square divisor sum. On the other hand, for odd composite $n$, square-free or not, 115 is the smallest of infinitely many odd composite $n$, none divisible by 3, whose divisor sums are square. This is shown in light of Dirichlet's theorem on the infinity of primes in any arithmetic progression $a + mb$, where $a$ and $b$ are co-prime.

If even $n = 2^m$, its divisor sum is not square for $m > 0$, but for $n$ containing any power of 2 and odd primes, square-free or not, 22 is the smallest of infinitely many even numbers whose divisor sums are square.

If $n$ is the cube of an odd prime $p$, it appears, with some help from triangle number theory, that $p^3$ is square (400) only for $p = 7$.

In all cases so far, the squares are even. But $n = p^4$, has square divisor sum for (and only for) $p = 3$. And the square is odd (121).

It is clear that the divisor sum must be even for odd powers of prime $p$, and odd for even powers of $p$. More generally, the divisor sum will be odd for $n$ containing $2^m$ (m > 0) and odd primes to even powers only. Odd divisor sums are a minority: the table through $n = 1000$ contains only 53 of them. 121 is the first square. And since 31 is the divisor sum of both 16 and 25, and for relatively prime integers, the sums are multiplicative, that is, the divisor sum of the product equals the product of the divisor sums of the factors, the divisor sum of 400 is a second odd square (961) in the table. 144 and 225 have divisor sum 403, but are not co-prime. Likewise for 242 and 196, with divisor sum 399. Going beyond the table, however, since 81 and 400 are relatively prime, the divisor sum of 81 x 400 = 32400 is $11^2$ x $31^2$ = $341^2$, a third odd square divisor sum.

Are there others? Or is it the case that 341, the smallest Fermat pseudoprime, is also the greatest odd number whose square is the divisor sum of an integer $n$? If so, is there a connection between odd square divisor sums and pseudoprimes?

• It's nice to have a general description, an introduction, ...--thank you. It'd be also nice to have a separately stated short question. Such a question would motivate reading the rest of the text well. – Włodzimierz Holsztyński Apr 3 '14 at 15:28

Searching OEIS for the first few terms gives A006532

From the comments there:

If a and b are in the sequence and relatively prime, then a*b is also in the sequence. - Franklin T. Adams-Watters, Jan 12 2009

Bunyakovsky's conjecture implies that this sequence is infinite, since then (e.g.) there are infinitely many primes of the form p = 3k^2 - 1, whence sigma(2p) = 3p + 3 = 9k^2. - Charles R Greathouse IV, May 12, 2011

Assuming infinitely many primes of the for $p=13x^2-1$, $\sigma(9 p)=13^2x^2$.

$341$ is not the largest prime with this property, check $9911$.

EDIT

For $\sigma(n)$ odd square the sequence is https://oeis.org/A008848

e.g. $\sigma(98551417041)=3^2 7^2 19^2 1093^2$

• But $13x^2-1$ can be prime only for even $x$, e.g. for $x = 6, 12, 18, 24, 30$, making $13^2x^2$ a divisor sum that is an even square. Can a divisor sum be an odd square $> 341^2$? – Edward Porcella Apr 3 '14 at 14:00
• @user48851 for sigma(n) odd square check: oeis.org/A008848 e.g. n=98551417041 – joro Apr 3 '14 at 14:09
• This answers the question, thank you – Edward Porcella Apr 4 '14 at 7:00