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For $n\in \mathbf{N}$ is $$n^3 \text{ divides } \sum_{i=1}^{n-1}\binom{n}{i}^2=\binom{n}{1}^2+\cdots +\binom{n}{n-1}^2$$ impling $$n \text{ divides } \sum_{i=1}^{n-1}\binom{n}{i}=\binom{n}{1}+\cdots +\binom{n}{n-1}$$?

Notice that the second condition is equivalent to say: "$n$ is a prime or $n$ is a Poulet number". Or

is $$n^4 \text{ divides } \sum_{i=1}^{n-1}\binom{n}{i}^2=\binom{n}{1}^2+\cdots +\binom{n}{n-1}^2$$ impling $$n \text{ divides } \sum_{i=1}^{n-1}\binom{n}{i}=\binom{n}{1}+\cdots +\binom{n}{n-1}$$? Notice that the first condition here is equvalent to say: "$n$ is a Wolstenholme number"

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    $\begingroup$ Please use $n$ instead of $p$. It is confusing as $p$ usually denotes prime numbers. $\endgroup$
    – Daniel Soltész
    Commented Feb 16, 2015 at 17:10
  • $\begingroup$ You might note the closed forms of the sums also: that is likely to get more people involved in answering. $\endgroup$
    – The Masked Avenger
    Commented Feb 16, 2015 at 17:29

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More of a comment. The first sum is $\binom{2n}{ n} - 2,$ the second sum is $2^n - 2.$ If $n$ is prime, then it divides the second sum by Fermat's little theorem, and divides the first sum by Wolstenholme's Theorem, as pointed out by the OP in the comment. In fact, the OP's question would follow from the Converse to Wolstenholme's Theorem.

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  • $\begingroup$ One reason to use these forms is that it may be possible to characterize those n for which gcd(n, 2n choose n) > 2. In particular, most composite integers n will not be expected to divide (2n choose n) - 2. $\endgroup$
    – The Masked Avenger
    Commented Feb 16, 2015 at 21:02

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