hello, see also here http://math.uci.edu/~tchoi/notes/wolstenholme.pdf Sonnhard.

Sayan wrote: Find all primes $p$ such that the expression \[\binom{p}1^2+\binom{p}2^2+\cdots+\binom{p}{p-1}^2\] is divisible by $p^3$. Isn't it true for any prime>3? Please check my proof: For any positive integer $i$ less than $p$, we know \[\frac{1}{p^2}\cdot \binom{p}{i}^2\equiv \frac {1} {i^2} \binom {p-1}{i-1}^2\equiv \frac {1}{i^2}\pmod p\] So we have to find those primes for which $\sum _{i=1}^{p-1}\frac 1 {i^2}$ is divisible by $p$. But it is obvious for any prime greater than $3$, since this sum is equivalent to $\sum _{i=1}^{p-1}i^2=p\cdot \frac {(p-1)(2p-1)}{6}$.

Yes, it seems that it does hold for all primes other than $2$ and $3$. Since $p|\binom{p}{k}$, $p^2|\sum_{k=1}^{p-1}\binom{p}{k}^2$. Now we need to show that $p|\frac{\sum_{k=1}^{p-1}\binom{p}{k}^2}{p^2}$. This follows from the fact that the above sum contains all quadratic residues modulo $p$ exactly twice so it is basically equal to $\sum_{i=1}^{p-1}i^2 = \frac{(p-1)(p)(2p-1)}{6}$ which is divisible by $p$.

dibyo_99 wrote: Yes, it seems that it does hold for all primes other than $2$ and $3$. Since $p|\binom{p}{k}$, $p^2|\sum_{k=1}^{p-1}\binom{p}{k}^2$. Now we need to show that $p|\frac{\sum_{k=1}^{p-1}\binom{p}{k}^2}{p^2}$. This follows from the fact that each term in the sum has a distinct residue modulo $p$ which is quite easy to prove. Well, my above proof shows exactly two terms of this sum have same residue.

Um well... yes. Sorry. I think we both know that I meant that the sum is the sum of all squares $(mod$ $p)$. Anyway, I'll edit it.

oh, it's nice problem! My solution same coincide with solution by PARTICLE. Answer: $p$ any prime numbers greater than $3$. Have someone more examples, which similar the problem?