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?