Recent content by MAGICMATHS

how can i explain that 6k is divisible by 6 when if i change it to asquared - 1 (pls refer to previous message) it becomes 36ksquared - 1...thats not easy to explain as divisible by 6!:(

1. My difficulty is to show that if a is an integer such that 2 does not divide a and 3 does not divide a then 24 does not divide a squared minus 1 2.Is there any equation which helps? 3. My idea is that it has to be an integer such that 6 does not divide a...therefore i have to show...

I don't know Fermat`s little Theory, i just started number theory to be honest ..is there a possibility using div alg pls?

I am trying to solve how an integer is simultaneously that is simultaneously a square and a cube number must be either of the form 7k or 7k+1 and I am failing when i work (7k+2)^2, even (7k+3)^2....Can i interpret 7k+1 as 7k+(-1), i think i can't but then i fail in many steps! Also i know that...

Thanks HallsofIvy, youve made my work very easy now. :)

Thanks DonAntonio, that's a very good clue which i did not think about. :)

How can i prove that if n>=1, (n(n+1)(2n+1))/6 is an integer. The hint is to use the division algorithm such that n has one of the forms 6k,6k+1,..6k+5 and to work each case...I tried changing n to 6k but i failed immeaditely :(