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The Fermat-Catalan conjecture is that $a^m + b^n = c^k$ has only a finite number of solutions when $a, b, c$ are positive coprime integers, and $m,n,k$ are positive integers satisfying $\frac{1}{m} + \frac{1}{n} +\frac{1}{k} <1$. There are currently only 10 solutions known (listed at the end of this post). My question concerns the case where $a, b, c$ are positive coprime Gaussian integers. I've found two solutions. Is there a clever method for finding more? I've used brute force techniques.

  • $(8+5i)^2+(5+3i)^3=(1+2i)^7$
  • $(20+9i)^2+(1+8i)^3=(1+i)^{15}$
  • $(1+2i)^7+(49+306i)^2=(27+37i)^3$ (Zander)
  • $(44+83i)^2+(31+39i)^3=(5+2i)^7$ (Zander)
  • $(238+72i)^3+(7+6i)^8=(7347−1240i)^2$ (Oleg567)

Here are the known solutions over integers.

  • $1^m+2^3=3^2$
  • $2^5+7^2=3^4$
  • $13^2+7^3=2^9$
  • $2^7+17^3=71^2$
  • $3^5+11^4=122^2$
  • $33^8+1549034^2=15613^3$
  • $1414^3+2213459^2=65^7$
  • $9262^3+15312283^2=113^7$
  • $17^7+76271^3=21063928^2$
  • $43^8+96222^3=30042907^2$
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I take it you're the E Pegg who submitted those two to mathworld? That's sort of cool. –  mixedmath Oct 2 '11 at 17:23
    
Is this you? –  Pedro Tamaroff Apr 30 '12 at 16:51
    
Yes, it's me. I'm the same guy. –  Ed Pegg May 1 '12 at 14:22
    
Hi Ed, I remember you from NKS SS 1. –  user02138 Jun 7 '12 at 21:35
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I can't contribute a clever method, but I also found these by brutish force: $(1+2i)^7+(49+306i)^2=(27+37i)^3$ and $(44+83i)^2+(31+39i)^3=(5+2i)^7$. –  Zander Jun 8 '12 at 2:25
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2 Answers

up vote 3 down vote accepted

I found the next complex solution! :) $$(238+72i)^3+(7+6i)^8=(7347−1240i)^2$$

(There is no new method here. Just small contribution to the problem.)

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Added it to the list. What were the bounds of your search? –  Ed Pegg Nov 13 '12 at 14:47
    
The bounds of my search were: $|\mathrm{Re}a|<B^{1/2}$, $|\mathrm{Im}a|<B^{1/2}$, $|\mathrm{Re}b|<B^{1/3}$, $|\mathrm{Im}b|<B^{1/3}$, $|\mathrm{Re}c|<B^{1/8}$, $|\mathrm{Im}c|<B^{1/8}$, where $B = 0\mathrm{x}80.00.00.00.00$ (hexadecimal) ~ $550\cdot 10^9$. Search form: $\pm a^2 \pm b^3 \pm c^8 =0$. –  Oleg567 Nov 13 '12 at 22:08
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Here is a small portion of what I have found:

$i^{4 m}+2^3=3^2$ where integer $m \ge 2$, i.e. the smallest is $i^8+2^3=3^2$

$(1+i)^2=i^{4 m+1}+i^{4 n+1}$ where integer $m \ge 1$, $n \ge 1$, i.e. the smallest is $(1+i)^2=i^5+i^5$

$(78-78 i)^2=(23 i)^3+i^{4 m+3}$ where integer $m \ge 1$, i.e. the smallest is $(78-78 i)^2=(23 i)^3+i^7$

$(11+11 i)^2+i^{4 m+1}=(3 i)^5$ where integer $m \ge 1$, i.e. the smallest is $(11+11 i)^2+i^5=(3 i)^5$

$(34-34 i)^2=(5 i)^3+(3 i)^7$

$(36+19 i)^2+(9-8 i)^3=(1+i)^{13}$

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+1 for last solution $(1+i)^{13}$ ! 2 examples more with $(ai)^n$: $$(3i)^{13} + (2761-2761i)^2 = (239i)^3,$$ $$(2797i)^3 + (13i)^9 = (75090-75090i)^2.$$ –  Oleg567 Dec 5 '12 at 13:17
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