Assume that q > p > 3. p and q are distinct odd primes. Prove that p > (3 + 6/(q-3)) and conclude that (p-1)(q-1) < n < (3(p-1)(q-1))/2
Explanation / Answer
given: q > p > 3, p and q are distinct odd primes since q can minimum value is 7 and the minimum value of p will be 5. so maximum value of 3+(6/(q-3)) will depend on minimum value of (q-3) and it implies that maximum value of 3+(6/(q-3)) depends on the minimum value of q which is 7. maximum value of 3+(6/(q-3)) =3+(6/(7-3))=4.5 maximum value of 3+(6/(q-3)) is 4.5 and it is less than the minimum value of p which is 5. hence, p > (3 + 6/(q-3)) (p-1) and (q-1) are an even numbers. (p-1)(q-1) will be even number and it will be divisible by 4 because each is an even number. from basic factor theorem, if x*y (+ve value) is divisible by 4 then at least one integer between x*y and (3/2)*x*y. hence, (p-1)(q-1)