# Why $\binom{mp-1}{p-1} \equiv 1 \pmod {p^3}$?

Why $$\binom{mp-1}{p-1} \equiv 1 \pmod {p^3}?$$

I tried by induction on $$m$$:

$$\bullet$$ $$m=0$$ $$\binom{-1}{p-1} = (-1)^{p-1}=1 \equiv 1 \pmod {p^3}$$

$$\bullet$$ $$m=1$$ $$\binom{p-1}{p-1} =1 \equiv 1 \pmod {p^3}$$

$$\bullet$$ induction hylpothesis: $$\binom{mp-1}{p-1} \equiv 1 \pmod {p^3}$$

$$\bullet$$ $$m+1$$ $$\binom{(m+1)p-1}{p-1}= \binom{(m+1)p}{p}-\binom{(m+1)p-1}{p}$$ but I am not able to move forward. Have you any hints?

Thank you so much

• Assuming $p$ is supposed to be prime, then I try $p=2, m=2$ and get $\binom 31=3\neq 1 \bmod 8$ – Mark Bennet Jan 27 at 16:54
• The $p \ge 5$, $m=2$ cases are Wolstenholme's theorem. (The primes $p=2$ and $p=3$ already don't work.) Experimentally, though, the result does seem true for all $p \ge 5$ and all $m$. – Misha Lavrov Jan 27 at 17:17

The necessary tools can be found in Glaisher's 1900 paper "Congruences relating to the sums of products of the first $$n$$ numbers and to other sums of $$n$$ products", which is more or less available here. (It is the first article in the issue.) Glaisher only proves the $$m=2$$ case this way, I think - but the method works for all cases.
Define the polynomial $$f(x) = (x+1)(x+2)\dotsb (x+p-1) = \left[{p\atop 1}\right] + \left[{p\atop 2}\right]x + \dots + \left[{p\atop p}\right]x^{p-1}$$ where $$\left[{p\atop k}\right]$$ denotes the unsigned Stirling number of the first kind. (Glaisher does not use this terminology or notation.)
The idea is that $$\binom{mp-1}{p-1}$$ can be written as the ratio $$\frac{f((m-1)p)}{f(0)}$$, and so it is only necessary to show that $$f((m-1)p) \equiv f(0) \pmod{p^3}$$ and also that neither is divisible by $$p$$, to prove your theorem. (Well, $$f(0)$$ is just $$(p-1)! = \left[{p\atop 1}\right]$$, so we know all about it, and it's not divisible by $$p$$.) This goes in $$3$$ steps:
1. We have $$f((m-1)p) \equiv \left[{p\atop 1}\right] + \left[{p\atop 2}\right]((m-1)p) + \left[{p\atop 3}\right]((m-1)p)^2 \pmod {p^3},$$ since all other terms have a factor of $$p^3$$ in them.
2. Actually, when $$p>2$$, $$\left[{p\atop 3}\right]$$ (as well as all other coefficients except $$\left[{p\atop 1}\right]$$) is divisible by $$p$$, so the quadratic term also vanishes, and we get $$f((m-1)p) \equiv \left[{p\atop 1}\right] + \left[{p\atop 2}\right]((m-1)p) \pmod {p^3}.$$ To see the divisibility by $$p$$, Glaisher compares the coefficients in $$f(x+1)$$ and $$(x+1)f(x)$$, and uses the known divisibility properties of binomial coefficients.
3. Actually, when $$p>3$$, $$\left[{p\atop 2}\right]$$ is divisible by $$p^2$$, so the linear term vanishes, and we get $$f((m-1)p) \equiv \left[{p\atop 1}\right] \pmod{p^3}.$$ To see this, Glaisher expands $$f(-p)=\left[{p\atop 1}\right]$$, and then all terms except the $$\left[{p\atop 2}\right]$$ term have a factor of $$p^2$$ in them.
This completes the proof, since we've concluded that $$f((m-1)p) \equiv f(0) \pmod{p^3}$$, and since neither is divisible by $$p$$, we can divide and get $$\frac{f((m-1)p)}{f(0)} \equiv 1 \pmod{p^3}$$.