According to Catalan, _{2N}C_{N} = ∑_{I} _{N}C_{I} ^{2}. Since _{N} C _{0} = _{N} C _{N} = 1, there’s where our 2 comes from. It turns out that _{N} C _{x} where 1 < X < N. Therefore N C _{I} = 2 + ∑ _{1}* _{N} C _{I} = 2 + (N * k) for some k. This means that the remainder will always be 2. As far as the converse, consider 343 as an example. _{686} C _{343} (mod 343) = 2, and yet 343=7 ^{3} and thus not prime.*

One issue with this: I have yet to find a counterexample that isn’t in the form p^{q} where p and q are both prime. Interesting?

*Cool problem huh? Anyone want to conjecture on that final form of the answer?*

## 1 comment:

Excellent puzzle. At the right level for my rusty brain.

Thoroughly enjoyed solving it, though it made me open my old probability book for definitions and properties of combinations. ( Its been years since I studied math. Too much software makes your brain soft. :-) )

No luck with the converse though. I'm going to try the first puzzle now.

Reg. mathml why can't you use something like this? http://pear.math.pitt.edu/mathzilla/itex2mmlFrag.html

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