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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....

Ryan gomez
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1 Answers1

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For $m>n>0$ , $$m^2-n^2=(m-n)(m+n)$$ is composite unless

  • $m-n=1$
  • $m+n=2n+1$ is prime
Peter
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  • so this is how i should write the proof to this question? – Ryan gomez Jul 13 '19 at 17:48
  • Actually, the answer to the question is "no" , so it is not a proof, but the description of the conditions for $m$ and $n$ to get a composite number. As you can see, there are counterexamples. – Peter Jul 13 '19 at 17:50