Dear friends and visitors of the site,

this is the April 1st problem:

Prove that for any prime number m > 3, m4 -1 is a multiple of 48.

Answer Submission Is Not Available

Form is valid through April 2018


Dear friends and visitors of the site.

Thank you for numerous answers and solutions.

The first to send a correct solution was Jacqueline  and we congratulate her!

Here is my solution to the problem:

m4 -1 = (m -1)(m+1)( m2 +1). m -1, m+1 and m2 +1 are even, so it is a multiple of 16 (one of them is a multiple of 4).

Between m-1, m+1 and m one is a multiple of 3, but not m.

Thus 16x3=48

Thank you and try to solve the May problem.

Serge Hazanov