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Does any row of Pascal's triangle contain a Pythagorean triple?

That is,

Are there integers $n,a,b,c$, where $0\leq a,b,c \leq n$, such that $\binom{n}{a}^2+\binom{n}{b}^2=\binom{n}{c}^2$ ?

Using $\binom{n}{r}=\frac{n!}{r!(n-r)!}$, this is equivalent to

$$[b!(n-b)!c!(n-c)!]^2+[a!(n-a)!c!(n-c)!]^2-[a!(n-a)!b!(n-b)!]^2=0$$

I used Excel to check if any of the first 141 rows contains consecutive entries $a,b,c$ such that $\binom{n}{a}^2+\binom{n}{b}^2=\binom{n}{c}^2$, and I did not find any. (Edit: Excel has limited precision, so it is unreliable for this question.) My question does not require that $a,b,c$ are consecutive entries in a row, but I do not know how to check all combinations of three numbers in each row.

Context: I sometimes ask questions about Pascal's triangle, such as this currently unresolved question about three consecutive integers.

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  • $\begingroup$ You can also write it as (a!(n-a)!)^(-2) + (b!(n-b)!)^(-2) = (c! (n-c)!)^(-2). So maybe the parametrization en.wikipedia.org/wiki/Fermat%27s_Last_Theorem#n_=_%E2%88%922 is useful. $\endgroup$
    – user146125
    Commented Sep 11 at 14:02
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    $\begingroup$ FWIW assuming without loss of generality that $0<a<b<c\leq\tfrac n2$ it follows that $$\binom ba^2+\binom{n-a}{n-b}^2=\left(\binom nc\frac{\binom ba}{\binom na}\right)^2.$$ $\endgroup$
    – Servaes
    Commented Sep 11 at 18:33
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    $\begingroup$ The answer is yes; one example has been found. I have asked a follow-up question at MO: How many other Pythagorean triples are contained in a single row of Pascal's triangle? $\endgroup$
    – Dan
    Commented Sep 12 at 4:31
  • $\begingroup$ Another follow-up question: "How many primitive Pythagorean triples are found in the interior of Pascal's triangle?" $\endgroup$
    – Dan
    Commented Sep 18 at 5:59

1 Answer 1

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Check:

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

I can share my Mathematica code if you want so. I am still running the program but so far this is the only solution.

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    $\begingroup$ I checked it by hand, and if you remove the shared factors it boils down to $3^2 + 4^2 = 5^2$. Really cool. $\endgroup$
    – user146125
    Commented Sep 11 at 15:52
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    $\begingroup$ I have asked a follow-up question at MO: How many other Pythagorean triples are contained in a single row of Pascal's triangle? $\endgroup$
    – Dan
    Commented Sep 12 at 4:32
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    $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Shaun ♦
    Commented Sep 12 at 17:11

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