# elem. # theo - "prove never perfect square"

• September 23rd 2008, 06:34 PM
cassiopeia1289
elem. # theo - "prove never perfect square"
Prove $3a^2 - 1$ is never a perfect square.
The book hints to use the fact: "The square of any integer is either of the form 3k or 3k+1," which we proved in a previous problem.

So how do you start?

Assuming $3a^2 - 1$ is a perfect square, it must be able to be written in the form of 3k or 3k+1.
Then what? How do you show it can't or NEVER can be?
• September 23rd 2008, 06:54 PM
Jhevon
Quote:

Originally Posted by cassiopeia1289
Prove $3a^2 - 1$ is never a perfect square.
The book hints to use the fact: "The square of any integer is either of the form 3k or 3k+1," which we proved in a previous problem.

So how do you start?

Assuming $3a^2 - 1$ is a perfect square, it must be able to be written in the form of 3k or 3k+1.
Then what? How do you show it can't or NEVER can be?

i guess you could assume a is even and show it doesn't work, then assume a is odd and show it doesn't work either. that is, you can't simplify it to get it in that form
• September 23rd 2008, 07:05 PM
cassiopeia1289
Ok, so by showing that there's no possible way to put it into the form would be like "let a=2s+1" ... ect ... which is not of the form 3k or 2k+1 and so on with "let a+2s" ... ? Is that enough?

But wait: how do we even know a is an integer at all? The problem never said so ... ?
• September 23rd 2008, 07:11 PM
Jhevon
Quote:

Originally Posted by cassiopeia1289
Ok, so by showing that there's no possible way to put it into the form would be like "let a=2s+1" ... ect ... which is not of the form 3k or 2k+1 and so on with "let a+2s" ... ? Is that enough?

But wait: how do we even know a is an integer at all? The problem never said so ... ?

yes, if you can show that, it is enough, because there are no other options. you have exhausted all possibilities. a is an integer, it has to be even or odd, and that covers all integers. so you would prove that it can never work no matter what
• September 23rd 2008, 08:54 PM
ThePerfectHacker
Quote:

Originally Posted by cassiopeia1289
Assuming $3a^2 - 1$ is a perfect square, it must be able to be written in the form of 3k or 3k+1.

Because $3a^2 - 1 = 3a^2 - 3 + 2 = 3(a^2 - 1)+2$.
Therefore it has the form $3k+2$ not $3k,3k+1$.