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This question leads to a follow-up: are there any Eisenstein triples (satisfying $a^2\pm ab+b^2=c^2$) in one row of Pascal's triangle apart from the following:

$\binom{23}{8}^2+\binom{23}{8}\binom{23}{9}+\binom{23}{9}^2=\binom{23}{10}^2$

In the referenced question a similar query is posed regarding Pythagorean triples for which the only apparent solution is

$\binom{62}{26}^2+\binom{62}{27}^2=\binom{62}{28}^2.$

Both the Pythagorean and Eisenstein solutions quoted above are embedded in a general set of three-term arithmetic sequence solutions, out of which only these two allow a triangle with the given numbers as sides and a rational-degree angle (a property held only by Pythagorean and Eisenstein triples).

We know from comments to the referenced question that no other Pytagorean solutions exist in any single row up to row $23000$. But, is there another Eisenstein solution in a single row, and also (since the known one has $a^2+ab+b^2=c^2$) is there any single-row solution at all using $a^2-ab+b^2=c^2$?

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  • $\begingroup$ if $a,b$ are consecutive in a Pascal's triangle row with $a^2 + ab + b^2 = c^2$ we do get things involving the next row, in that $a^2 - a (a+b) + (a+b)^2 = c^2 $ $\endgroup$
    – Will Jagy
    Commented Sep 18 at 19:10
  • $\begingroup$ Yes indeed. But can we do that in one row? I modified the q slightly to bring that out. $\endgroup$ Commented Sep 18 at 19:14

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