Re: Special primes

In message <hi68e11jo98d32juujo8bcodv9qnnennn6@xxxxxxx>, quasi <quasi@xxxxxxxx> writes 
On Sun, 24 Jul 2005 15:02:34 EDT, bischar <bisch_a_r@xxxxxxxx> wrote:

For every prime numbers m and n, with n and m consecutive and n<m, we have m²-n² ending with only 0, 2 or 8.

The only digit ending two consecutive 'm²-n²' is 0... and if you suppress the numbers 'm²-n²' with 0 as last digit you see a suite 'm²-n²' ending with 2, 8, 2, 8, 2, 8, 2, 8, never two 2 or 8 consecutively... (verified for 5<m<10000).

Funny no ?

Do you see a logic or a reason for that ?

Can you verify it automatically for greater primes ? 

A fascinating pattern.

The pattern is true for 5<n<10^6.

There may be an easy explanation but I don't see it.

quasi


The last digits of primes >5 fall into two pairs, (1,9) and (3,7) (depending on the last digits of their squares). If m^2 - n^2 is 0 then m and n have last digits from the same pair. If it's 2 then n has last digit 3 or 7 and m has 1 or 9, and if its 8 vice versa.

As n increases, each non-zero in the sequence of last digits of m^2 - n^2 corresponds to a change in which pair the last digit of n is from, and so must alternate between 2 and 8.

This applies to any increasing sequence of odd numbers which are not multiples of 5, not just primes.
--
David Hartley
.

Relevant Pages
Quantcast