Having trouble proving these two statements? I have some type of idea but not specifics.

Question

Prove: If $n$ is in the set of integers, then $4|n^2$ or $4|(n^2-1)$.

My Thoughts: I thought to use proof by cases because of the disjunction “or”. Case 1 would’ve been even and the Case 2 would’ve odd. When using the case of even, both n^2 and (n^2 -1 ) show that they are divisible by 4. But for odd cases does not work. Not sure what to do?

Prove: The product of any “n” consecutive positive integers is divisible by n!. My thoughts- proof by contradiction & that $n!=n(n-1)(n-2)(n-3)...(n-k)!$.

Thanks a bunch.

Answer 1

Case 1. $n$ is even. Then $n=2k$ where $k$ is integer. It follows that $4$ divides $n^2$ because $n^2=(2k)^2=4k^2$.

Case 2. $n$ is odd. Then $n=2k+1$ where $k$ is integer. In that case, $4$ divides $n^2-1$ since $n^2-1=(2k+1)^2-1=4k^2+4k=4(k^2+k)$.

Note that $n^2$ and $n^2-1$ can't be both divisible by $4$ since they differ by $1$.

Answer 2

Hint for 1: consider $(2k)^2 \space and \space (2k - 1)^2$

Hint for 2: not so easy, two alternative approaches:

1. Try to prove that power of prime p in n! is not bigger than power of p in a product of n consecutive positive integers.

2. Search and read about Binomial coefficients