number theory

any integer may be the form 2k or 2k+1 so let m=2k and n=2p+1 then nm(n-m) = (2p+1).2k.(2p+1-2k) which is even i totalwe have 4 different cases and in case the exp. is even

yup.. this can work

Otherwise write the expression as nm(n-m) = nm [(n-1)-(m-1)] = nm(n-1) - nm(m-1)

Since k(k-1) is always even the above expression is also always even

Thanks for the replies...

2)If, (x+y)/2 = (y+z)/3=(z+x)/4, then find x:y:z