Question

Prove that there exist n consecutive positive integers each having a (nontrivial) square factor. How would you then modify your proof so that each of these integers instead has a cube factor (or more generally, a kth power factor where k ≥ 2)?

This is a number theory question.
Please show all steps and make clear notes about what is happening for a clear understanding.
Please write clearly or do in latex.
Thank you

Answer 1

ANSWER:

Assume k and n exist in the positive integers and that k is less than n. If k is less than n, then

n!=1⋅2⋅3⋅⋯⋅k⋅⋯⋅n,

which is to say that kth is a factor of n!. So,

k+n!=k+(1⋅2⋅⋯⋅k⋅⋯⋅n)

=k(1⋅2⋅⋯⋅n/k+1)

Remember that k is a factor of n! and so n!/k is still an integer.

So since k is an integer that is bounded between 11 and n, it stands that whatever number you pick up to nn can divide k+n!k+n! making it composite till the nth integer, but n!n! has that nth integer in it so the nth integer is also composite which means that you can pick any integer between 1 and n inclusively and it will be composite.

since we can say that there exist n consecutive postive integers each having a square factor.

(Thank You).