If you already have no idea what I’m talking about; please read the last posts.

Let with . Then we have if, and only if ;

And this is equal to if, and only if . And this suffices to prove the theorem I mentioned in the previous post. And, by Wolstenholme´s theorem, we know that for all . So for every number , for which holds, we have that . Assume for the moment that the following converse holds: is the only for which . While this is not true (, is a counterexample), there should be a relatively low upper bound for the amount of with and . In any case, bear with me for the moment. Let’s pick some ‘random’ number and let’s assume that it lies pretty much in the middle of two consecutive powers of . Then, the amount of numbers , with and for which , is roughly equal to . Now the sweet part: if we sum over all primes , then what we obtain is not more than the number of such that is divisible by *any *prime . And . So at least about 25% of all the with is not divisible by any prime , which implies .

Obviously, this doesn’t come very close to a real proof. But it might be a good heurstic start. Anyway, it certainly seems necessary to have some bounds on the for a given , such that and . Let me know what you think!

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