May $p^3$ divide $(a+b)^p-a^p-b^p$? Do there exist positive integers $a,b$ and a prime $p>\max(a,b)$ such that $p^3$ divides $(a+b)^p-a^p-b^p$? 
The reader of Kvant magazine A. T. Kurgansky asked to prove that such $a,b,p$ do not exist, see here.
But discussion here 
On the exact reference of a cute Diophantine problem
suggests that it should be very hard to prove. Maybe, a counterexample may be bound? Roughly speaking, a probability of this event is about $1/p^2$, for each $p$ we have about $p$ events (even for $b=1$, I was previously wrong that it may be supposed without loss of generality, thanks for Noam Elkies for noting this) and so as $\sum 1/p=\infty$, we may expect them (and even infinitely many!) But this series converges very slowly, so the minimal example may be large.
 A: We explain the pattern observed by Joe Silverman, deducing
the existence of infinitely many solutions, some of which even have
$p^5 | (a+b)^p - a^p - b^p$.
Lemma. If $n \equiv 1 \bmod 3$ then the homogeneous polynomial
$P_n(a,b) := (a+b)^n - a^n - b^n \in {\bf Z}[a,b]$ has a factor $(a^2+ab+b^2)^2$.
Proof :  Either
i) evaluate $P_n(x,1)$ and its derivative at a cube root of unity $\rho$, or
ii) use the $S_3$ symmetry of $-P_n(a,b) = a^n + b^n + c^n$ where $a+b+c=0$:
double roots at $(a:b:c) = (1:\rho:\rho^2)$ and $(1:\rho^2:\rho)$
are the only ways to get the total number of roots to be $1 \bmod 3$. $\Box$
[These ${\bf Q}(\rho)$-rational points on the $n$-th Fermat curve 
are well known.]
Corollary. If $p$ is a prime congruent to $1 \bmod 3$,
and $a,b$ are integers such that $p^k | a^2+ab+b^2$, then
$p^{2k+1} | P_n(a,b)$.
Proof : Observe that $P_p \in p {\bf Z}[a,b]$ and use the Lemma.
Now $k=1$, and thus $2k+1=3$, is easy to obtain:
choose $a$ arbitrarily, and let $b\equiv ra \bmod p$ where $r$ is a cube root of unity
mod $p$.  We can even get a factor of $p^5$ by choosing $a,b$ such that
$a^2 + ab + b^2 = p^2$ (that is, so that $a,b,p$ are sides of
a triangle with a $120^\circ$ angle); for example,
$$
(3+5)^7 - 3^7 - 5^7 = 120 \cdot 7^5.
$$
A: There are apparently lots of examples.  The smallest is $a=1$, $b=2$, and $p=7$.
A: For what it's worth, consider $b=p-1$. It seems (experimentally at least, I haven't tried to prove it) that for every $p\equiv1\pmod6$ there exists an $a$ in the range $2\le a\le \frac{1}{2}p$ such that 
$$(a+p-1)^p-a^p-(p-1)^p\equiv0\pmod{p^3}.\qquad (*)$$
I checked for $p<500$, and in this range, for each $p$ there is a unique such $a$.
Further, again just for $p<500$, if $p\not\equiv1\pmod6$, then there are no solutions to $(*)$.
