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Solution: Day 2, problem 1

The units form a cyclic group <a∣a5=1>< a | a^{5} = 1 >. Since -1 is a unit of order dividing 2, it equals 1, so 2 = 0. A computation then shows that the element x=1+a2+a3x = 1 + a^{2} + a^{3} has cube equal to 1.
Since 3 is prime to 5, we must have x=1x = 1, i.e. a2(a+1)=0a^{2}(a + 1) = 0. But then a+1=0a + 1 = 0, which is a contradiction.