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

andn-k is 4 or larger, there are no solutions.” I was taking the interpretation “If k is 4 or larger (or equivalentlyif 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.