5
$\begingroup$

Solutions to the equation $$\dbinom{a}{n}+\dbinom{b}{n}=\dbinom{c}{n}$$ I will refer to as 'Binomial triplets of order $n$'. These triplets describe simplicial $n$-polytopic numbers that can be written as the sum of two $n$-polytopic numbers.

The above equation has infinitely many solutions for $n=1$ (trivial) and $n=2$ (the triplets $(a,b,c) =$ $(3,3,4)$, $(4,6,7)$, $(6,7,9)$, $(5,10,11)$, $(7,10,12)$, etc.). Also, regardless the value of $n$, it allows for at least one solution $(2n-1,2n-1,2n)$.

Is it true that for any $n$ the above equations has infinitely many solutions? Are there infinitely many $n$th order Binomial triplets?

[Edit]It appears this problem is known as Bombieri's Napkin Problem.[\Edit]

$\endgroup$
  • 2
    $\begingroup$ For $n=3$, this is $f(a)+f(b)=f(c)$ for some particular cubic polynomial $f$. I'd expect such an equation to have only finitely many solutions, and ditto for any $n>3$. Note $${10\choose3}+{16\choose3}={17\choose3}$$ $\endgroup$ – Gerry Myerson Aug 16 '14 at 12:50
  • $\begingroup$ Two more: $\binom{22}{3}+\binom{56}{3}=\binom{57}{3}$ and $\binom{32}{3}+\binom{57}{3}=\binom{60}{3}$. $\endgroup$ – Johannes Aug 17 '14 at 14:15
  • $\begingroup$ OEIS has listed quite a few tetrahedral numbers that are the sum of two other tetrahedral numbers: oeis.org/A034404 . $\endgroup$ – Johannes Aug 17 '14 at 14:41
  • $\begingroup$ So, I knew a lot more about this problem 4 years ago than I knew yesterday. Anyway, it seems that what is known about the question is at the MO link. $\endgroup$ – Gerry Myerson Aug 18 '14 at 1:40
1
$\begingroup$

Some computations were reported by John Leech, Some solutions of Diophantine equations, Math Proc Camb Phil Soc 53 (1957) 778-780. For $n=3$, he reported 17 solutions with $c<500$. For $n=4$ and $c<500$ he found only $(a,b,c)=(190,132,200)$ (and the trivial $(7,7,8)$). For $n\ge5$, he found only the trivial $(2n-1,2n-1,2n)$ (it's not clear to me what range of values of $n$ and of $c$ were tested).

Computers have come a long way since 1957. I'm sure someone could push this farther (and maybe someone already has). But of course that won't answer the questions about whether there are an infinity of solutions.

Some other solutions are reported at http://www.numericana.com/fame/apery.htm One infinite family begins $(n,a,b,c)=(6,19,19,21),(35,118,118,120),(204,695,695,697),\dots$. Another begins $(n,a,b,c)=(6,14,15,16),(40,103,104,105),(273,713,714,715),\dots$. This at least shows there are nontrivial solutions for arbitrarily large $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.