BdMO National Higher Secondary 2007/4

[BdMO]

Problem 4:
$(m,n)$ represents the largest common divisor of integers $m$ and $n$. For example $(2,3)=1$ and $(10,15)=5$. Suppose $n(n+1)(n+2)$ is a square, where $n$=integer.
a) What is $(n,n+1)$?
b) What is $(n+1,n+2)$?
c) What is $(n+1,n(n+2))$?

From your answers $a,b,c$ is it possible for $n(n+1)(n+2)$ to be a square?

[Cryptic.shohag]

n, (n+1), (n+2) are back to back integers. If n is even, then (n+1) is odd and (n+2) is even. And if n is odd, then (n+1) is even and (n+2) is odd. So, (n+1) is coprime to both n and (n+2). So....

a) (n,n+1)=1
b) (n+1,n+2)=1
c) (n+1, n(n+2))=1

Now, as there is no comon factor between (n+1) and n(n+2), n(n+1)(n+2) cannot be a square....

[Masum]

There is no common factor between $2^2,3^2$ but their product is $6^2$

[Cryptic.shohag]

There is no common factor between $2^2,3^2$ but their product is $6^2$

$2^2,3^2$ are not back to back integers but n,(n+1) are. Putting square makes difference here....

[Moon]

Now, as there is no comon factor between (n+1) and n(n+2), n(n+1)(n+2) cannot be a square....

Actually Masum wanted to say about this logic. When you use some logic it should be true in all cases. I mean you said that $gcd(n,n+1)=1 \Longrightarrow n(n+1)(n+2)$ is not a square, your logic sound something like $gcd(a,b)=1 \Longrightarrow ab(\text{something relatively prime to b})$ can not be a square.

And your solution has a few problems (that are fixable easily, though). The line I quoted is not correct in all cases.

[Cryptic.shohag]

Actually my logic was if a and b r back to back common integers then $(a,b)=1$ and ab can't be a square. Is there any problem in my logic moon vaia? If so, plz give me an example....