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At MSE I asked, "Does any row of Pascal's triangle contain a Pythagorean triple?" The answer is yes; the example $\binom{62}{26}^2+\binom{62}{27}^2=\binom{62}{28}^2$ was given. In that answer's comments (which have been moved to chat), the answerer reports that this is the only solution found in the first $5199$ rows.

Now my question is,

How many other Pythagorean triples are contained in a single row of Pascal's triangle?

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    $\begingroup$ I have checked all binomials up to $n$=3300 and so far managed to find only the solution that is already quoted. $\endgroup$
    – Oldboy
    Commented Sep 12 at 7:38
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    $\begingroup$ This paper shows that $n=62,k=26$ is the only solution of the form $\binom{n}{k}^2+\binom{n}{k+1}^2=\binom{n}{k+2}^2$. $\endgroup$
    – N M
    Commented Sep 12 at 8:02
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    $\begingroup$ We do have $\binom{23}{8}^2+\binom{23}{8}\binom{23}{9}+\binom{23}{9}^2=\binom{23}{10}^2$ in which the named binomial coefficients are the legs and hypotenuse of a $120°$-angle triangle. $\endgroup$
    – Oscar Lanzi
    Commented Sep 12 at 15:57
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    $\begingroup$ I have to update my previous comment (maybe for someone who wants to start his/her own computation): $\binom{n}{a}^2+\binom{n}{b}^2=\binom{n}{c}^2$ has no other solution for $n<23000$ $\endgroup$
    – Oldboy
    Commented Sep 18 at 7:48
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    $\begingroup$ I decided to stop my program. There is no solution up to $n=30000$. The code is now dealing with such a huge numbers that it's highly unlikely that any three of them will make a match. $\endgroup$
    – Oldboy
    Commented Sep 26 at 8:22

1 Answer 1

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The known Pythagorean solution is actually part of a larger pattern in which three consecutive elements appear in arithmetic progression in some row. To wit, if the chosen arithmetic progression is proportional to $k-2,k,k+2$ then the elements

$a=\binom{k^2-2}{\frac12[k(k-1)]-2},b=\binom{k^2-2}{\frac12[k(k-1)]-1},c=\binom{k^2-2}{\frac12[k(k-1)]}$

satisfy the condition $a:b:c=(k-2):k:(k+2)$.

The only positive three-term arithmetic progressions that are also Pythagorean triples are, of course, $\{3,4,5\}$ and its multiples, so only the $k=8$ solution above (using $2$ as the common difference) is found by this method.

However, we also have the progression $\{3,5,7\} (k=5)$, in which the numbers represent the sides of an obtuse triangle with a $120°$ angle. Such triples (Eisenstein triples) satisfy the relation $a^2+ab+b^2=c^2$, and so

$\binom{23}{8}:\binom{23}{9}:\binom{23}{10}=3:5:7$

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

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