Sum of divisors mod 8

Question

I am wondering if anyone knows if there are any formulas or results about the sums of divisors of an odd perfect square mod 4,8, 12 etc or anything helpful?

Answer 1

  n: 1  n^2: 1  sigma(n^2) 1 mod eight:   1
  n: 3  n^2: 9  sigma(n^2) 13 mod eight:   5
  n: 5  n^2: 25  sigma(n^2) 31 mod eight:   7
  n: 7  n^2: 49  sigma(n^2) 57 mod eight:   1
  n: 9  n^2: 81  sigma(n^2) 121 mod eight:   1
  n: 11  n^2: 121  sigma(n^2) 133 mod eight:   5
  n: 13  n^2: 169  sigma(n^2) 183 mod eight:   7
  n: 15  n^2: 225  sigma(n^2) 403 mod eight:   3
  n: 17  n^2: 289  sigma(n^2) 307 mod eight:   3
  n: 19  n^2: 361  sigma(n^2) 381 mod eight:   5
  n: 21  n^2: 441  sigma(n^2) 741 mod eight:   5
  n: 23  n^2: 529  sigma(n^2) 553 mod eight:   1
  n: 25  n^2: 625  sigma(n^2) 781 mod eight:   5
  n: 27  n^2: 729  sigma(n^2) 1093 mod eight:   5
  n: 29  n^2: 841  sigma(n^2) 871 mod eight:   7
  n: 31  n^2: 961  sigma(n^2) 993 mod eight:   1
  n: 33  n^2: 1089  sigma(n^2) 1729 mod eight:   1
  n: 35  n^2: 1225  sigma(n^2) 1767 mod eight:   7
  n: 37  n^2: 1369  sigma(n^2) 1407 mod eight:   7
  n: 39  n^2: 1521  sigma(n^2) 2379 mod eight:   3
  n: 41  n^2: 1681  sigma(n^2) 1723 mod eight:   3
  n: 43  n^2: 1849  sigma(n^2) 1893 mod eight:   5
  n: 45  n^2: 2025  sigma(n^2) 3751 mod eight:   7
  n: 47  n^2: 2209  sigma(n^2) 2257 mod eight:   1
  n: 49  n^2: 2401  sigma(n^2) 2801 mod eight:   1
  n: 51  n^2: 2601  sigma(n^2) 3991 mod eight:   7
  n: 53  n^2: 2809  sigma(n^2) 2863 mod eight:   7
  n: 55  n^2: 3025  sigma(n^2) 4123 mod eight:   3
  n: 57  n^2: 3249  sigma(n^2) 4953 mod eight:   1
  n: 59  n^2: 3481  sigma(n^2) 3541 mod eight:   5
  n: 61  n^2: 3721  sigma(n^2) 3783 mod eight:   7
  n: 63  n^2: 3969  sigma(n^2) 6897 mod eight:   1
  n: 65  n^2: 4225  sigma(n^2) 5673 mod eight:   1

Answer 2

Let $\sigma(x)$= the sum of the divisors of x. It can be shown that If gcd(x,y)=1, then $\sigma(xy)=\sigma(x)\sigma(y)$.

By the fundamental theorem of arithmetic, $x=p_1^{e_1}p_2^{e_2}...$. Powers of primes are relatively prime. The sum of powers of primes is a geometric series.

$\sigma(p_1^{e_1})=1+p_1+p_1^2+p_1^3+...+p_1^{e_1}$

Every prime greater than 2 is one more or less than a power of 4.

So: $\sigma(p_1^{e_1})=e_1+1 (mod 4)$ if $p_1=1(mod 4)$

If $p_1=-1(mod 4)$. We know $e_1$ is even since we are dealing with a perfect square.

So $\sigma(p_1^{e_1})=1. $

So $\sigma(n^2)=(-1)^k$ where $k$ is the number of exponents in the prime factorization which are even and not divisible by 4 and associated with primes 1 more than a multiple of 4.

Every prime greater than 5 is 1,-1, 3, or 5 more than a multiple of 8.

$3^2=9$ and $5^2=25$ are both one more than a multiple of 8.

If $p_1=1 (mod 8)$ then $\sigma(p_1^{e_1})=e_1+1(mod 8)$

if $p_1=-1(mod 8)$ then $\sigma(p_1^{e_1})=1(mod 8)$

if $p_1=3(mod 8)$ then $\sigma(p_1^{e_1})=2e_1+1(mod 8)$.

if $p_1=5(mod 8)$ then $\sigma(p_1^{e_1})=3e_1+1(mod 8)$

You can imply something similar to 12.