Binomial coefficients are hardly ever powers. That is, there are strong restrictions on when the equation
has integer solutions for l ≥ 2.
- There are infinitely many solutions for k = 1 or 2.
- If k = 3 and l = 2, there is only one solution: n = 50 and m = 140.
- If k is 4 or larger, there are no solutions.
(Binomial coefficients remain unchanged when you replace k with n–k. So when we say there are no solutions for k ≥ 4, we also assume n-k ≥ 4.)
Source: Proofs from the Book
I must be mis-understanding something. Suppose we re-write every instance of k as (n-k):
I’m confused how there can be infinitely many solutions with k=1 or k=2 and at the same time (n-k) is 3 or less. Those constraints only permit a handful of (n-choose-k) pairs: (3-choose-1)=3, (2-choose-1)=2, (5-choose-2)=10, (4-choose-2)=6, and (3-choose-2)=3… none of which is of the form m^l with l>1?
F. Carr: I’m not sure if I understand what you’re asking, so I’ll give an example.
Suppose n = 100. Then the theorem says that C(100, k) could only possibly be an integer power (with exponent at least 2) if k = 1, 2, 98, or 99, leaving aside the trivial cases k = 0 and k = 100. As it turns out, C(100, 2) and C(100, 98) are not powers. But C(n, 2) is a square for infinitely many values of n.
Ah. I think the statement
is intended to be interpreted as “If k is 4 or larger and n-k is 4 or larger, there are no solutions.” I was taking the interpretation “If k is 4 or larger (or equivalently if n-k is 4 or larger), there are no solutions.” Thanks, now it is clear to me.
What about k=3 and l > 2?
For fixed l, this leads to a curve of genus > 1 for which we want to know the integer points, so we now that their number is finite by theorems by Siegel etc..
For general l, I suppose we could use the ABC theorem (as soon as it’s confirmed) to know that there’s a finite total number of solutions.