Perfect Gaussian

[Ed Pegg]

About 30 years ago, I asked on mathpuzzle.com if there was a perfect Gaussian integer.

Today, dimitrovich702 sent me 

2 (430089 + 665198 I) == Total[Divisors[430089 + 665198 I]]  

Seems to check out.  Is this significant in any way?  


I also learned of a Spira/Ward/McDaniel definition that parallels classic perfect numbers.  

p = 73;  m = (1 + I)^p - 1;  z = -I (1 + I)^(p - 1) m;

{DivisorSigma[1, z] == (1 + I) z, z}

{True, 4722366482869645213696 - 4722366482800925736960 I}  

73 starts the list {73, 113, 241, 353, 457, 3041, 27529, 364289, 991961, 1203793, 1667321, 4792057}.   
Members of https://oeis.org/A057429 == 1 (mod 8)

[Allan Wechsler]

I have dabbled in this stuff, and discovered that there is a lot of hidden nuance in how you define DivisorSigma. Because I'm not a fluent Mathematica user, I found the description on the Wolfram reference page not easily understandable.

As soon as you start to talk about Gaussian integers, the first question you must answer is, "Is 5 perfect?" After all, according to an obvious possible definition, the Gaussian divisors of 5 are 1, 2+i, 2-i, and 5. Their sum is 10. 10 is twice 5.

If 5 is not "Gaussianly" perfect according to your definition, you need to be really explicit about definitions.

-- Allan

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[DONG HAOXUAN]

Ed, Allan,

Allan's point about the definition turned out to be important. I
looked into this more carefully.

Let S(z) denote the sum obtained from Wolfram/Mathematica's Gaussian
Divisors convention, where one representative from each associate
class is chosen with positive real part and nonnegative imaginary
part. Under this convention, the original example is indeed correct:

z = 430089 + 665198 i

satisfies

S(z) = 2z.

I factored it as

z = -(1+2i)^3 (5+2i) (38+17i) (316+5i),

reconstructed all 32 divisor representatives, and their sum is exactly

860178 + 1330396 i = 2z.

There is also a stronger result. After some theoretical reductions,
all remaining cases below this norm can be reduced to a finite
exhaustive search using exact arithmetic. The exhaustive search was
then independently reimplemented, and both implementations give

N(z) < 627464927125 => S(z) != 2z.

At N(z) = 627464927125, all 32 possible Gaussian associate classes
were checked directly, and exactly one normalized solution occurs:

430089 + 665198 i.

Thus, under the Wolfram Divisors convention, this is the unique
normalized solution of minimum norm, and therefore also of minimum
absolute value. The minimality result is a computer-assisted proof
rather than a purely hand proof; the exhaustive search has been
independently reimplemented and reproduced.

There is an important distinction from the classical Spira/McDaniel
notion of a perfect Gaussian integer. The direct sum of Wolfram's
first-quadrant divisor representatives is not multiplicative and
depends on the representative convention, whereas the Gaussian
DivisorSigma/Spira definition uses a multiplicative choice of
representatives arising from Gaussian prime factorization. In
particular, the number above is not perfect in the Spira sense.

So I think there are really two separate questions here: the classical
multiplicative notion of Gaussian perfection, and the
convention-dependent arithmetic problem S(z) = 2z arising from
Mathematica's Divisors representatives.

I have the exact search code and logs if anyone would like to check them.

Best,
Jason Dong

On Wed, 19 Aug 2026 18:44:57 -0400, Allan Wechsler <acw...@gmail.com> wrote:
> I have dabbled in this stuff, and discovered that there is a lot of hidden nuance in how you define DivisorSigma. Because I'm not a fluent Mathematica user, I found the description on the Wolfram reference page not easily understandable.
>
> As soon as you start to talk about Gaussian integers, the first question you must answer is, "Is 5 perfect?" After all, according to an obvious possible definition, the Gaussian divisors of 5 are 1, 2+i, 2-i, and 5. Their sum is 10. 10 is twice 5.
>
> If 5 is not "Gaussianly" perfect according to your definition, you need to be really explicit about definitions.
>
> -- Allan
>
> On Sun, Aug 16, 2026 at 11:03 AM Ed Pegg <edp...@gmail.com> wrote:
>

> About 30 years ago, I asked on [mathpuzzle.com](http://mathpuzzle.com) if there was a perfect Gaussian integer.

>
> Today, dimitrovich702 sent me
>
> 2 (430089 + 665198 I) == Total[Divisors[430089 + 665198 I]]
>
> Seems to check out. Is this significant in any way?
>
> I also learned of a
>
> Spira/Ward/McDaniel definition that parallels classic perfect numbers.
>
> p = 73; m = (1 + I)^p - 1; z = -I (1 + I)^(p - 1) m;
>
> {DivisorSigma[1, z] == (1 + I) z, z}
>
> {True, 4722366482869645213696 - 4722366482800925736960 I}
>
> 73 starts the list {73, 113, 241, 353, 457, 3041, 27529, 364289, 991961, 1203793, 1667321, 4792057}.
> Members of https://oeis.org/A057429 == 1 (mod 8)
>
> --
>
> You received this message because you are subscribed to the Google Groups "SeqFan" group.
>
> To unsubscribe from this group and stop receiving emails from it, send an email to seqfan+un...@googlegroups.com.
>

> To view this discussion visit [https://groups.google.com/d/msgid/seqfan/353e08b3-9a69-4ca4-b302-07be76f2d947n%40googlegroups.com](https://groups.google.com/d/msgid/seqfan/353e08b3-9a69-4ca4-b302-07be76f2d947n%40googlegroups.com?utm_medium=email&utm_source=footer).

>
> --
>
> You received this message because you are subscribed to the Google Groups "SeqFan" group.
>
> To unsubscribe from this group and stop receiving emails from it, send an email to seqfan+un...@googlegroups.com.
>

> To view this discussion visit [https://groups.google.com/d/msgid/seqfan/CADy-sGHwjioZg-muxKX0iNmzLsQV71X8KRWAcQxshLt%2B-octiw%40mail.gmail.com](https://groups.google.com/d/msgid/seqfan/CADy-sGHwjioZg-muxKX0iNmzLsQV71X8KRWAcQxshLt%2B-octiw%40mail.gmail.com?utm_medium=email&utm_source=footer).