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Jul 11, 2015 at 1:15 comment added Marty Isaacs @Ladisch--Right. I should have thought of that.
Jul 10, 2015 at 22:34 comment added Frieder Ladisch This is a very nice argument, and it seems to me that it works for all $q>2$: just replace $-I$ by $aI$ where $a\in \mathbf{F}_q^*$ has prime order $p$ (multiplicatively, so $p$ divides $q-1$).
Jul 10, 2015 at 17:57 history answered Marty Isaacs CC BY-SA 3.0