N and X are constant. p,a and n change but the invariant holds at the beginning and end of every iteration.

Proof by induction: the invariant is true at the start when X=a, N=n and p=1. If n is even we have a<-a^2 and n<-n/2 and the invariant holds. If n is odd, a^n*p becomes a^(n-1)*(a*p) which is equal to a^n*p and the invariant holds. Thus the invariant holds in both cases.

I must confess I spent almost a whole night thinking about it. If you interested in knowing more about invariants, checkout "Discipline of Programming" Edsger Dijkstra or "Science of Programming" David Gries.

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bjam â€“with-test variant=release link=shared address-model=64

should be:

bjam â€“-with-test variant=release link=shared address-model=64

b.regards!

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