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Prove $\displaystyle 3a^2 - 1$ 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 $\displaystyle 3a^2 - 1$ 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 formQuote:
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 whatQuote:
Originally Posted by cassiopeia1289
Because $\displaystyle 3a^2 - 1 = 3a^2 - 3 + 2 = 3(a^2 - 1)+2$.Quote:
Originally Posted by cassiopeia1289
Therefore it has the form $\displaystyle 3k+2$ not $\displaystyle 3k,3k+1$.
