In 1823, the Academy of Science in Paris offered a prize for a
solution to Fermat's
Last Thereom. Despite the many proofs they received, the first gold
medal they presented was to Ernst
Kummer, who looked at complex number
sets he called the ideals.
Now, Wiles has settled FLT for integers. But what about FLT for Gaussian
Integers? For that matter, how about Beal's Conjecture
or the ABC
Conjecture -- How do the Gaussian Integers fit in?

**Prize Problem -- $500**. Claim: For an integer *n* >
2,
there are no solutions in the Gaussian integers to the equation x^{n}+y^{n}=z^{n}.
I hereby offer $500 for a counterexample, with requirements |x y z|
> 0 and {x,y,z} Gaussian
Integers. (I've no interest in impossibility
proofs). Dave Rusin pointed me to a math-atlas note.W.
Edwin Clark pointed me to this article.

**Prize Problem -- $50**. Claim: Beal's Conjecture
is true for Gaussian integers as well. I offer $50 for a proper
counterexample. Solved by
Fred W. Helenius: (-2+i)^3 + (-2-i)^3 = (1+i)^4. This was the
only
example he found. I'll offer $10 for another (and I should have
found that one).

Prize Problem -- $10. Two
positive Gaussian Integers, a
and b, are chosen at random.
What is the exact probability that GCD[a,b]
= 1? If a and b are plain integers, the answer is
6/Pi^{2}, as noted in the Mathworld
GCD entry. Solved in
1988 by George Collins. W. Edwin Clarke pointed me to the original paper.

Perfect
numbers. 6, 28, and 496 have the property that they are the sum of
their divisors. Gaussian Integers also have divisors. The
divisors of 5 are {1, i, -1, -i, 1+2i, -2+i, -1-2i, 2-i, 2+i, -1+2i,
-2-i, 1-2i, 5, 5i, -5, -5i}. It's always possible to factors a
Gaussian Integer into positive Gaussian Primes of the form x+yi with
x>0 and y>=0, and one of the four roots of unity. For example, 5
= -i * 1+2i * 2+i. When Divisors are considered, only the "positive"
Gaussian integers are listed.

Other than itself, the positive divisors of 3185+2912i are 1, 2+3i,
3+2i, 5+12i, 7, 13, 13+2i, 14+21i, 20+43i, 21+14i, 35+32i, 35+84i,
39+26i, 41+166i, 91, 91+14i, 140+301i, 169+26i, 245+224i, 273+182i,
287+1162i, 455+416i, and 1183+182i. The sum of these positive divisors
of 3185+2912i happens to be 3183+ 2912i. Other close misses include
13+16i, 227+364i, 319+458i, 513+284i, 516+313i, 313+516i, 637+780i,
896+553i, 1401+938i, 938+1401i, 1853+1254i, and 2224+1867i. I
offer a $100 prize to the
first
to find a proper Gaussian perfect number. W. Edwin Clark pointed
me to this article.

The numbers 220 and 284 are amicable, because each is the sum of the
divisors of the other. Are there amicable Gaussian Integers? My small
search didn't find any. I didn't find any Aliquot cycles, either.
I offer a $50 prize to
the
first to find a proper Gaussian Aliquot cycle.

Any aliquot
sequence can terminate with 0, grow without bound to infinity, or
go
into a cycle. One aliquot cycle of length 5, found in 1918 by
Poulet, is 12496, 14288, 15472, 14536, and 14264. 2856 eventually
leads to a 28-cycle. Perfect numbers are length 1 cycles, Amicable
numbers are length 2 cycles, Sociable numbers are length 3+ cycles.

A Sierpinski
Number is a number k such that k 2^{n} + 1 is always
composite. I discovered that 10+3i is a Sierpinski number.

Mersenne Primes: Mike Oakes has been investigating the Gaussian
Mersenne numbers since 1968. Prime
Number Group, Mail
Archive, Integer
Sequences Gaussian Mersenne.

Euler's Formula, n^2 + n + 41. Is there a GP equivalent?
The formulas n^2 - n + (9+4i) and n^2 + n + (9+4i) both seem to
be a very rich prime finders for a+bi such that (a+b)<21. I
have no idea why (9+4i) is special.

It is an unsolved problem whether there are infinitely many primes of
the form a^2 + 1. Whether there are infinitely many primes of
the form (a+bi)^2 + 1 is easy to figure out (why?). Less easy is
(a+bi)^n + gcd(a,b)+1. There seems to be an infinite number of primes
of this form for any n. For (8+i)^n + 2, the expression is prime for
n=1,2,3,4,5.

The Catalan
Conjecture has been solved for normal integers. Fred W.
Helenius showed that there are several solutions among the Gaussians.
(78+78i)^2 + (23i)^3 = i, 1
+ (1-i)^5 = (1+2i)^2, i
+ (11+11i)^2 = (3i)^5.

The Fermat-Catalan
conjecture looks for high powers adding to other powers. 1^{p}
+ 2^{3} = 3^{2} (*p* __>__ 2), 2^{5}
+ 7^{2}
= 3^{4}, 13^{2} + 7^{3} = 2^{9}, 2^{7}
+ 17^{3} = 71^{2}, 3^{5} + 11^{3}
=
122^{2}, 33^{8} + 1549034^{2} = 15613^{3},
1414^{3} + 2213459^{2} = 65^{7},
9262^{3}
+ 15312283^{2} = 113^{7}, 17^{7} + 76271^{3}
= 21063928^{2}, and 43^{8} + 96222^{3} =
30042907^{2} are the known examples. In Gaussian Integers, I
found (8+5i)^{2} + (5+3i)^{3} = (1+2i)^{7},
(20+9i)^{2}
+ (1+8i)^{3} = (1+i)^{15}. Are there more? I
offer $10 for each new
example,
or each new class of Gaussian Fermat-Catalan numbers. New
examples
should all be relatively prime with each other.

Fred W. Helenius found 5: (5i)^3 + (3i)^7 = (34-34i)^2, (49+306i)^2 +
(1+2i)^7 = (27+37i)^3, (44+83i)^2
+ (31+39i)^3 = (5+2i)^7, (19+36i)^2
+ (1-i)^13 = (9+8i)^3, (2+i)^4
+ (1+i)^9 = (5+4i)^2.

Thokchom Sarojkumar Singh sent some examples where all parts had a
common factor, but it could not be removed to obtain a different
solution. I'm not sure what to call these: (1 - 3I)^5 + (1
- 18I)^2 = (1 + 2I)^4, (1 + 3I)^4 + (1 + I)^13 = (8 - 10I)^2, (3
+ 2I)^3 + (1 - 2I)^3 = (4 + 6I)^2, (3 + 5I)^4 + (1 - 4I)^5 = (31 -
22I)^2, (33 + 6I)^2 + (6 - 8I)^3 = (1 + 2I)^6, (1 + 2I)^6 + (3 + I)^7 =
(25 + 50I)^2, (1 + 5I)^5 + (29 + 2I)^2 = (1 - 8I)^4 .

(3+13i)^{3} + (7+i)^{3} = (3+10i)^{3} +
(1+10i)^{3}
and (6+3i)^{4} + (2+6i)^{4} = (4+2i)^{4} + (2+i)^{4}.
Is there something similar for Gaussian fifth powers? See http://euler.free.fr/details.htm
for details on this problem in the integers. Solved by Fred W. Helenius:
(2+3i)^5 + (2-3i)^5 = 3^5 + 1,
(1+6i)^5 + (3-2i)^5 = (6+i)^5 + (-2+3i)^5, (9+6i)^5 + (3-10i)^5 =
(6+i)^5 + (6-5i)^5,
(15+14i)^5 + (5-18i)^5 = (18-7i)^5 + (2+3i)^5.

In doing these computer searches, keep in mind the words of D.H.
Lehmer: "Happiness is just around the corner."

P. Poulet, #4865, L'intermediare des math. 25 (1918), pp. 100-101.

I wrote a notebook to show a bit
of what *The
Mathematical Explorer* can do. Here is GaussMeetsFermat.nb.
And here is an
HTML version of the notebook. I found a neat image (below). The
entire notebook was built in *The Mathematical Explorer*
(though I did use *Mathematica* to generate HTML). Voronoi
diagrams.

*a+bI* connected to Round[(*a+bI*)^(1/3)]^3 with The Mathematical
Explorer