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This question relates to the class of solutions of the Diophantine equation:

$$a^3+b^3+c^3=d^3$$

meeting all the following conditions: $a,b,c,d>0$; $gcd(a,b,c,d)=1$; two odd and two even terms; $d$ odd.

Such a solution must have $a+b+c$ odd, implying $a+b+c \equiv 1$ or $3 \pmod4$. The twelve smallest solutions (ranked by value of $d$) show a surprising pattern in respect of $a+b+c\pmod4$:

$$1^3+6^3+8^3=9^3\qquad1+6+8=15\equiv3\pmod4$$ $$3^3+10^3+18^3=19^3\qquad3+10+18=31\equiv3\pmod4$$ $$4^3+17^3+22^3=25^3\qquad4+17+22=43\equiv3\pmod4$$ $$2^3+17^3+40^3=41^3\qquad2+17+40=59\equiv3\pmod4$$ $$6^3+32^3+33^3=41^3\qquad6+32+33=71\equiv3\pmod4$$ $$29^3+34^3+44^3=53^3\qquad29+34+44=107\equiv3\pmod4$$ $$22^3+51^3+54^3=67^3\qquad22+51+54=127\equiv3\pmod4$$ $$36^3+38^3+61^3=69^3\qquad36+38+61=135\equiv3\pmod4$$ $$14^3+23^3+70^3=71^3\qquad14+23+70=107\equiv3\pmod4$$ $$38^3+43^3+66^3=75^3\qquad38+43+66=147\equiv3\pmod4$$ $$25^3+48^3+74^3=81^3\qquad25+48+74=147\equiv3\pmod4$$ $$50^3+61^3+64^3=85^3\qquad50+61+64=175\equiv3\pmod4$$ The pattern does not continue: we have: $$20^3+54^3+79^3=87^3\qquad20+54+79=153\equiv1\pmod4$$ In fact, for solutions with $87\leq d\leq199$ there is the sort of random behaviour that might be expected: $17$ of $32$ solutions have $a+b+c\equiv1\pmod 4$ and the remaining $15$ have $a+b+c\equiv3\pmod4$.

Question Is there a reason why all solutions with $d<87$ have $a+b+c\equiv3\pmod4$, or is this just a surprising outcome of random behaviour?

Adam Bailey
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  • There's a parametrization by Euler (http://personal.psu.edu/lxv1/preprints/Rachel11.ps) but I haven't been able to extract anything from it yet. I did a quick computer search and found that 92% of the first 5868 solutions leave residue 3, so what you found is just due to the distribution, not a special feature of the first 12 solutions. Of course I don't know why the solutions are distributed this way. – Sophie Jan 21 '20 at 07:32
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    @Sophie I could not confirm your 92% finding. My method was to download from the website of Wroblewski the first of his (3,1,3) files which contains all solutions of $a^3+b^3+c^3=d^3$ with $a,b,c,d>0$, $gcd(a,b,c,d)=1$, and $d<100000$, import the file to Excel, and then set up suitable count functions over those solutions with two odd and two even terms and $d$ odd. For the first 5868 solutions (implying $d\leq 8641$) I found 49.9% residue 1 and 50.1% residue 3. For the whole range to $d=99999$ I found 49.7% residue 1 and 50.3% residue 3. – Adam Bailey Feb 07 '20 at 22:17

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