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Consider the binomial coefficients, excluding those of the form $\binom{n}{1}$ and $\binom{n}{n-1}$.

  • There are some that differ by $1$, for example $\binom{7}{2}-\binom{6}{3}=21-20=1$.
  • There are some that differ by $3$, for example $\binom{31}{2}-\binom{11}{5}=465-462=3$.
  • There are some that differ by $4$, for example $\binom{5}{2}-\binom{4}{2}=10-6=4$.
  • There are some that differ by $5$, for example $\binom{6}{2}-\binom{5}{2}=15-10=5$.
  • There are some that differ by $6$, for example $\binom{9}{3}-\binom{13}{2}=84-78=6$.
    Etc.

But I cannot find any that differ by $2$. I checked the first $120$ rows of Pascal's triangle. So, is the following conjecture true or false:

There are no binomial coefficients that differ by $2$, excluding those of the form $\binom{n}{1}$ and $\binom{n}{n-1}$.

Call it the "Twin binomial coefficient conjecture", if you will.

Context: I was playing with Pascal's triangle, trying to approximate the median of the numbers in the first $n$ rows, and I noticed a lack of differences of $2$.

Dan
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2 Answers2

18

Unfortunately this is not true, the smallest counterexample per list in A006987 is $$ \binom{604}{2}-\binom{104}{3}=2. $$

Sil
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  • 3
    Funnily enough, that's the only counterexample of that list. I wonder what the next smallest counterexample is. – Dan Nov 20 '23 at 12:25
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    At least finding examples of the form ${y \choose 2} - {x \choose 3} = \pm 2$ amounts to finding integer points on $2$ elliptic curves (one for each of $\pm$), and the one in this answer turns out to be the only nontrivial one. – Travis Willse Nov 21 '23 at 18:14
9

For the sequence ${{n + 1} \choose 2} = \frac{1}{2} n (n + 1)$, applying the backward difference operator gives $\Delta {{n + 1} \choose 2} = {{n + 1} \choose 2} - {n \choose 2} = n$, so there is no nontrivial solution of ${v \choose 2} - {u \choose 2} = \pm 2$.

Hence we next check for solutions of the form $${v \choose 2} - {u \choose 3} = \pm 2.$$ For the moment taking $+$, writing the left-hand side as a polynomial (and scaling by a factor of $6$ and rearranging) gives the elliptic curve $$3 v^2 - 3 v = u^3 - 3 u^2 - 2 u - 12 ,$$ and our problem is to locate all of the integer points on the curve. Already a general result about elliptic curves tells us that there are at most finitely many such points. Changing variables $u = \frac{x}{3}$, $v = \frac{y}{9}$ (and scaling) puts the curve in Weierstrass form with integer coefficients: $$y^2 - 9 y = x^3 - 9 x^2 + 18 x - 324 .$$

Then running the following SAGE routine finds all $8$ integral points on the curve:

R2.<x,y> = PolynomialRing(QQ,2)
E = EllipticCurve(y^2 - 9*y - (x^3 - 9*x^2 + 18*x + 324))
E.integral_points(both_signs=true)

Only $3$ of the points have positive $x$- and $y$-values, $(9 : 27 : 1)$, $(12 : 36 : 1)$, $(312 : 5436 : 1)$, and they respectively correspond via our change of variables to the solutions $$(u, v) = (3, 3), (4, 4), (104, 604).$$ (If you allow negative values $n$ in binomial coefficients $n \choose k$, the remaining $5$ integral points yield further solutions, but none are interesting.)

The first two solutions are trivial, and the last one (which is mentioned in Sil's answer) provides a counterexample to the conjecture, $$\boxed{\left({104 \choose 3}, {604 \choose 2}\right) = (182104, 182106)} ,$$ which is thus the only one of the form ${v \choose 2} - {u \choose 3} = 2$.

The $-$ case is similar, but the resulting equation admits no integer solutions (in fact, no rational solutions).

Travis Willse
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