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**alunw** This is clearly all very laborious but we could now complete the multiplication table for a field of 16 elements. Then we would have to (and could) show that x^2+x+1000 did no have any root in our field of 16 elements, and posit 10000 as one of its two roots, at which point we would know that 10000^2 = 11000. Then the same kind of reasoning would presumably eventually lead us to realise that 10*10000 and 100*10000 and 1000*10000 were all tricky products to calculate. Probably the best way to proceed would be to calculate all the powers of 10000 find its order (which presumably is a factor of 255 and probably is 255 though I don't know for sure - there certainly is an element of order 255 somewhere in the multiplicative group of non-zero elements in that field and very likely some structure theorem tells us that this is one of them).