if m>n>0, is m^2 - n^2 composite?
p is composite if p>1 and there exists positive integers r & s such that p = rs where 1< r < p and 1
m^2 - n^2= (m+n)(m-n) let m =2 and n =1 let p = m^2 - n^2 p = 3 or (3)(1) let r = (m+n) and s = (m-n) r satisfies that 1< r < p but s doesn't, so m^2 - n^2 is not composite
let m=4 and n =2 p = 12 or (6)(2) r satisfies that 1< r < p and s satisfies that 1
i am confused about this, how come there can be 2 possible answers to this question? if i answer yes, im right but also wrong....
For $m>n>0$ , $$m^2-n^2=(m-n)(m+n)$$ is composite unless