This isn't necessarily true. We simply want it to be so. Especially for m-1 and m-3, there are cases where both are divisible by two. However, we then know that m-2 and m-4 are coprime and could replace m-1 and m-3. Nevertheless, this first idea does not work since we cannot handle the term (m-1)^2 (or (m-2)^2). To ensure that a and a-3 are coprime, we select a such that it is not divisible by 3. Consequently, we achieve gcd(a,a-3) = gcd(a,3) = 1.
True. I intended to say that IF n satisfies the given condition, THEN m >= 11 as 9, 10 are consecutive divisors of n! Therefore, your example shows that n=6 doesn't satisfy the given condition.
I think a solution without Bertrand can be motivated from last year's imo p1
Sorry sir, can you explain for me the reason why both m-1 and m-3 must be co- prime similiary for (a-3) and a must be co- prime. I still don't get it.
This isn't necessarily true. We simply want it to be so. Especially for m-1 and m-3, there are cases where both are divisible by two. However, we then know that m-2 and m-4 are coprime and could replace m-1 and m-3. Nevertheless, this first idea does not work since we cannot handle the term (m-1)^2 (or (m-2)^2). To ensure that a and a-3 are coprime, we select a such that it is not divisible by 3. Consequently, we achieve gcd(a,a-3) = gcd(a,3) = 1.
@@calimath6701 ok thanks for the explaination
For n=6, m=7 is not >=11?
True. I intended to say that IF n satisfies the given condition, THEN m >= 11 as 9, 10 are consecutive divisors of n!
Therefore, your example shows that n=6 doesn't satisfy the given condition.