Complex number ABC Conjecture

Question

Regarding the abc conjecture, I can't find records for Gaussian integers.

Let $rad(a) = \lVert{\prod{prime factors(a)}}\rVert$.
For relatively prime $(a,b,c)$ with $a+b=c$, define quality as $Q = \frac{\log{c}}{\log(rad(a b c))}$.

For Gaussian integers, the highest quality I can find in initial searching is

$Q = 1.65169:\ -1 +\ (2 + i)^4 =\ i(1+2 i)(1+i)^7 : $
$Q = 1.50515:\ 1 +\ (2 + i)^2 =\ -(1+i)^5 : $
$Q = 1.34527:\ 1 +\ (8 + 7 i)^2 =\ -i(1+2 i)^2(1+i)^9 : $ $Q = 1.23754:\ -3^4 +\ (4 + i)^4 =\ -i(1+2 i)(2+ i)^2(1+i)^9 : $
$Q = 1.21850:\ (1+2 i)^2 +\ (4+i)^4 =\ -(1+i)^2 (3+2 i)^2 (2+i)^3 : $ $Q = 1.21684:\ -1 +\ (4+i)^4 =\ -i(1+i)^8(1+2 i)(2+i)(2+3 i) : $

Can anyone beat that first item, or supply other complex a+b=c sums with $Q>1.22$?

Answer 1

For equation $a+b=c$ $$i(1+i)\;+\;(2+3i)^4\;=\;i(3+2i)^4$$ we have $$Q = \frac{\log\left(13^2\right)}{\log\left(13\sqrt{2}\right)}\approx 1.761929.$$

Answer 2

My search strategy is to look at $p^x + uq^y$ for $u \in \{\pm 1, \pm i\}$, $x \le 8$, $y \le 8$ where $p$ and $q$ are Gaussian primes, $q$ comes before $p$ in my generation order, and (to reduce duplicates) $\mathfrak{Re}(p) \ge 0$, $\mathfrak{Im}(p) \ge 0$.

I've excluded the values which are already in the question or Oleg's answer. C# code available here.

$$\begin{matrix} Q=1.22286:& (1+i)^{6}(2-i)^{2}(4+i)(5+2i)^{2} - (3+2i)^4 &=& (4-i)^6 \\ Q=1.22315:& i(1+i)^{2}239 - (3+2i)^8 &=& (3-2i)^8 \\ Q=1.22581:& i(2+i)(13+10i)^{2} - i(1+i)^6 &=& (3-2i)^5 \\ Q=1.23422:& (2-i)^3 + (1+i)^4 &=& -(2+i)^{3} \\ Q=1.23453:& (2-i)^4 - i*(1+i)^{9} &=& (5-4i)^2 \\ Q=1.24114:& (1+i)^{8}(2+i)^{5}(2-i)^{5}19(5+4i)(5-4i) + 3^4 &=& 79^4 \\ Q=1.24243:& -(1+i)^{3}(3-2i)^{2} + i(2+i)^4 &=& (2-i)^5 \\ Q=1.24265:& 3^{3}7^{2} + i(1+i)^6 &=& 11^3 \\ Q=1.24394:& (6-i)^3 + i(1+i)^5 &=& i(2-i)^{5}(4+i) \\ Q=1.24405:& (1+i)(2+i)^{5} - i(3-2i) &=& -3^4 \\ Q=1.24417:& 11^2 + 7^5 &=& -i(1+i)^{10}23^{2} \\ Q=1.24509:& (1+i)^{5}3(2+i)^{4}7(4-i)^{2} - (3+2i)^4 &=& (6-i)^6 \\ Q=1.24669:& (4-i) + (4+i)^5 &=& i(1+i)^{6}(10+7i)^{2} \\ Q=1.24745:& (5-2i)^2 - (3+2i)^6 &=& (1+i)^{7}(2+i)^{4}(6-5i) \\ Q=1.25188:& (1+i)^{5}(2+i)^{2}(2-i)^{4}(4+i)(5-4i) - (5-2i)^6 &=& -(3+2i)^8 \\ Q=1.25770:& (5-4i)^5 + (2+i)^6 &=& -i(1+i)^{6}(3-2i)(4+i)^{3}(5-2i) \\ Q=1.25875:& (5-4i)^8 - (5+4i)^8 &=& -i(1+i)^{12}3^{2}(2+i)(2-i)^{1}7^{2}31 \\ Q=1.26345:& (4-i)^4 - (3-2i)^8 &=& -i(1+i)^{8}(2+i)^{3}(2-i)^{2}(4+i)(6-5i) \\ Q=1.26393:& (1+i)^{10}3^{2}(2+i)(4+i)(8-7i) + (2-i)^8 &=& (3-2i)^8 \\ Q=1.26423:& (4-i) + i(2+i)^8 &=& -(1+i)^{4}(11-6i)^{2} \\ Q=1.26731:& 3^4 - i(1+i)^{9}(2+i)(2-i)^{2} &=& (4-i)^4 \\ Q=1.26732:& (2-i)^4 - (2+i)^4 &=& -i(1+i)^{8}3 \\ Q=1.27108:& i(1+i)^{6}(2+i)^{4}(2-i)^{4}(17+8i)(17-8i) + 3^8 &=& 11^6 \\ Q=1.27587:& (3-2i)^8 + (1+i)^2 &=& (2-i)(3+2i)^{4}(5-2i)(14+i) \\ Q=1.27688:& (1+i)^{5}(3+2i)^{2}(4+i)^{2} - (2+i)^6 &=& (6-i)^4 \\ Q=1.28146:& (1+i)^{12}3^{2}(2+i)(2-i)(5+4i)^{2}(5-4i)^{2} - 31^4 &=& -7^8 \\ Q=1.29113:& i(3-2i)^5 - (1+i)3^{4}(2+i) &=& (5-2i)^4 \\ Q=1.29588:& 7^2 - i(1+i)^{10} &=& 3^4 \\ Q=1.30191:& (2-i)^8 - (2+i)^8 &=& (1+i)^{10}3 \times 7 \\ Q=1.31000:& (1+i)^{11}(2+i)^{2}(5+4i) + (2-i)^8 &=& (5-4i)^4 \\ Q=1.31721:& (1+i)3 - (2+i)^5 &=& i(2-i)^5 \\ Q=1.31933:& 3^4 - i(1+i)^{2}(2+i)^{4}(2-i)^{4} &=& 11^3 \\ Q=1.32040:& (4-i)^8 - (2+i)^4 &=& i(1+i)^{7}(2-i)^{3}(3-2i)^{2}(5+2i)(8-5i) \\ Q=1.32546:& i(1+i)(3-2i)^{4} + (4-i) &=& 3^5 \\ Q=1.34858:& i(1+i)^{14}3^{2}11 + 7^6 &=& 19^4 \\ Q=1.34933:& 23^3 + 11^2 &=& (1+i)^{24}3 \\ Q=1.36272:& (5-2i)^5 - i(2-i) &=& i(1+i)^{5}3^{2}(8-5i)^{2} \\ Q=1.37890:& 31^7 + 19^8 &=& i(1+i)^{22}7^{4}(93+20i)(93-20i) \\ Q=1.44082:& 83^5 + 7^2 &=& -(1+i)^{4}3^{12}(4+i)(4-i)(10+3i)(10-3i) \\ Q=1.49814:& i3(2-i)^{7} - (1+i) &=& (8-5i)^3 \\ Q=1.52038:& 1 + i(1+i)^6 &=& 3^2 \\ Q=1.69084:& 3^{10}(10+3i)(10-3i) - i(1+i)^2 &=& 23^5\\ \end{matrix}$$

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A variant approach using products of up to two prime powers (with no limit on the exponent) visited in order of magnitude by means of a priority queue quickly turns up some new values including the record (below, because I found it in an external source first) and

$$\begin{matrix} Q=1.22274:& (1+i)^{20} - 1 &=& -(2+i)^{2}(2-i)^{2}(5+4i)(5-4i) \\ Q=1.22641:& 3^{2}7 + 1 &=& -(1+i)^{12} \\ Q=1.23622:& -(12+7i) + i(1+i)^{18} &=& (3+2i)^{3}(2+i)^{3} \\ Q=1.24287:& (2+i)^{2}(1+i)^{14} + 1 &=& i3(2-i)^{4}(8-3i) \\ Q=1.24767:& (1+i)^{11}(2-i) - (2+i)^{8} &=& (8+3i)^{2}3^{2} \\ Q=1.25351:& i + i(4-i)^{2}(1+i) &=& (2+i)^{4} \\ Q=1.26010:& i(2+i)(2-i)^{3} - i(1+i)^{11} &=& (4+i)^{3} \\ Q=1.26186:& (3-2i)^{4}3 + i(1+i)^{13} &=& i(2+i)^{6}(4-i) \\ Q=1.26259:& (2-i)(5-2i) + i(3-2i)^{4} &=& (1+i)^{15} \\ Q=1.28053:& (1+i)^{2}(3-2i)^{4}(4+i) - i(6+i) &=& i(2+i)^{9} \\ Q=1.28228:& (5-4i)^{2}(2-i)^{3} + (1+i)^{13} &=& -3^{5}(2+i) \\ Q=1.28428:& i*(3+2i)^{2}(3-2i)^{2} + i*7^{3} &=& (1+i)^{18} \\ Q=1.28741:& 3^{2}(1+i)^{12} - 7^{2} &=& -(2+i)^{4}(2-i)^{4} \\ Q=1.31297:& (5+2i)^{3}(6-5i) + (2-i)^{9} &=& i(2+i)^{8}(1+i)^{3} \\ Q=1.31500:& (2+i)^{9} + (1+i)^{12} &=& (2-i)^{3}(3-2i)^{2}(9+4i) \\ Q=1.32443:& 3^{2}(1+i)^{10} + i &=& i(4+i)^{2}(4-i)^{2} \\ Q=1.34239:& (4+i)^{4} + (1+i)^{11} &=& -i(2+i)^{5}(5+2i) \\ Q=1.34514:& 3(2+i) - (5+4i)^{4} &=& (4-i)^{5}(1+i) \\ Q=1.35088:& (3+2i) - (18-5i)^{2}(1+i)^{4} &=& i(2+i)^{9} \\ Q=1.38279:& (3+2i)(3-2i) + 3^{5} &=& (1+i)^{16} \\ Q=1.38712:& (2+i)(2-i) - (1+i)^{20} &=& 7^{3}3 \\ Q=1.42160:& (1+i)^{18} + i &=& i*3^{3}19 \\ Q=1.42928:& (1+i)^{16} + i(7+2i)(2-i)^{2} &=& -(2+i)^{7} \\ Q=1.42934:& 1 - i11^{2}(1+i)^{2} &=& 3^{5} \\ Q=1.43862:& (1+i)^{8}(2+i)(2-i) + 1 &=& 3^{4} \\ Q=1.44012:& (2+i)^{8} - 1 &=& -(1+i)^{9}3(2-i)(4+i) \\ Q=1.44731:& (4-i)(2-i)^{6} + (4+i)(2+i)^{6} &=& (1+i)^{20} \\ Q=1.49092:& (1+i)^{9} + i(3-2i)^{3}(2+i) &=& -(2-i)^{6} \\ Q=1.58842:& (2+i)^{3}(2-i)^{3} + 3 &=& i(1+i)^{14} \\ \end{matrix}$$

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I've also found a Master's paper which presents some values, including in particular the record-breaking

$$Q=2: 1 + 1 = -i(1+i)^2$$