I’ve just realized a rather obvious fact concerning the vague question at the end of my latest post on Kloostermania, which I’ll rephrase informally:

Can one guess with better than even chance that a graph like Kloostermania’s represents a Kloosterman or a Salié sum?

I had said that, for fixed *p*, just knowing the sum shouldn’t leave much room to make any choice except a random one. To make sense, the rule must be made more precise: you are given a bare real number, say

and you are told that it is either the value of a Kloosterman sum *S(1,1;p)* for some prime, or of a Salié sum *T(1,-1;p)* for some prime. You can win a drink of your choice by picking up the right one. What can you say? The reason it is hard to do better than heads-Kloosterman/tail-Salié is that you do not know *p*. If you knew the value of the prime, then you could compute the angle in [0,π] such that

and use the fact that these are supposed to be distributed differently for Kloosterman and Salié sums. For instance, the probability that *θ* is(conjecturally) larger than 3π/4 is bigger for Salié sums (it is 1/4) than for Kloosterman sums (about 0.09…), and hence, finding an angle in this range would give a big hint that it comes from a Salié sum (but no certainty, of course).

And now for the really obvious point: if you can, in addition to knowing *S*, actually *see* the graph of the partial sums, then — of course — *you know the prime*: you just have to count the steps on the graph.

It seems that similar ideas should lead to a slightly better than average guess even without knowing *p*: the size of *S* gives a lower bound on *p*, and for each possible guess of *p*, we have a guess of either Salié or Kloosterman. Presumably, combining these should be possible to get a small gain on pure chance…

(Note: readers are welcome to make a guess concerning the value above…)

If one is given the pair (S,p), and one is also allowed to assume that the a priori chance of a Kloosterman or Salié sum occurring is 1/2, then it would seem that the obvious guess (determined by whether theta was in [pi/4,3 pi/4] or not) would work 1/2 + 1/(2*Pi) of the time, and would be the best guest.