Given $a,b,c \in \mathbb{N}$ which satisfy the following conditions:
$a^3 + b^3 = c^2$
$ a \neq b$
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EDIT, Will Jagy: The conjecture is that, for a given $c,$ there are at most two distinct pairs $(a,b)$ with $1 \leq a < b$ and $a^3 + b^3 = c^2.$ Note the example $77976^2=1026^3+1710^3=228^3+1824^3$ that Gerry found with exactly two pairs.
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There are maximum $2$ possible pairs of $(a, b)$ which satisfy this conditions.
I have verified it for $a, b \leq 200\,000$.
Note: the situation when $a$ and $b$ just take each others value is discarded.
According to oeis, the first square that can be expressed as a sum of two cubes in three different ways is $$3343221000^2 = 279300^3 + 2234400^3 = 790020^3 + 2202480^3 = 1256850^3 + 2094750^3.$$ Perhaps the smallest in four ways, $$\eqalign{42794271007595289^2 &= 14385864402^3 + 122279847417^3\cr &= 55172161278^3 + 118485773289^3\cr &= 64117642953^3 + 116169722214^3\cr &= 96704977369^3 + 97504192058^3\cr}$$ Possibly the smallest in five ways: $$\eqalign{47155572445935012696000^2 &= 94405759361550^3 + 1305070263601650^3 \cr&= 374224408544280^3 + 1294899176535720^3 \cr&= 727959282778000^3 + 1224915311765600^3 \cr&= 857010857812200^3 + 1168192425418200^3 \cr&= 1009237516560000^3 + 1061381454915600^3\cr}$$