Mike Hansen <mhan...@gmail.com> writes: >> I'm not sure whether you saw my answer yet... It shows that you can have >> full evaluation (as in Python), and still work modulo n. > > William was just saying that the mod function in > mod(2^(2^517)+1,84977118993*2^520+1) couldn't easily recognize of the > structure in the arguments since they are evaluated before mod sees > them. > >> (I'd be surprised and disappointed if sage cannot do this.) > > Here is your example in Sage: > > sage: F = Integers(84977118993*2^520+1) > sage: F(2)^(2^517)+1 > 0 > sage: F(2)^(2^516)+1 > 207830575673500686411447362595659393768941593937702880049854622986883156049131962664562951494983277006774963177214890291645066756995999343954972963541296782130435362337
precisely! thanks. BTW: could you alternatively specify the expected return type? I.e., tell sage: please use those operations that make the result be in F? Martin -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
