Again, will be one of the factors of (i hope by now you see why. if not, think about it some more)Now show that m divides n! + m, for all m such that 2<m<n (here it is a less than or equal to sign) and
by the above, it would be divisible by . hence, it cannot be prime, it must be composite.show why n! +m is composite.
take the sequenceUse this to find a sequence of n-1 consecutive integers (that is a sequence a2, a3,...,an, with ai+1=ai +1 and all ai composite).
do you see why this works?