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 xⁿ+yⁿ=zⁿ. 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/Pi², 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 posssible 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 2ⁿ + 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 + 2³ = 3² (p > 2), 2⁵ + 7² = 3⁴, 13² + 7³ = 2⁹, 2⁷ + 17³ = 71²,  3⁵ + 11³ = 122²,  33⁸ + 1549034² = 15613³,  1414³ + 2213459² = 65⁷,  9262³ + 15312283² = 113⁷,  17⁷ + 76271³ = 21063928²,  and 43⁸ + 96222³ = 30042907² 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