Prove that 3n^2 - 1 can't be a square of a integer n

In summary, the conversation discusses the problem statement of proving that 3n^2 - 1 cannot be the square of an integer, with the method of assuming the opposite and reaching a contradiction. A suggestion is given to consider remainders mod 4 as a possible approach, but the other person clarifies that modulus has not yet been studied.
  • #1
walker242
12
0
Well, the problem statement is in the title:
Given that n is an integer, show that 3n2 - 1 can't be the square of an integer.

Currently, I don't have any idea at all where to start. Method is probably to assume opposite and show that this leads to a contradiction.

Any hint as to where to start would be very appreciated!
 
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  • #2
Look at remainders mod 4. What are the possible values for n^2 mod 4?
 
  • #3
Thanks for the reply!

While that probably is one way of looking at the problem, we haven't yet reached modulus in our studies.
 
  • #4
You don't really need to study modulus to think about remainders after division by 4. Nothing else comes to mind as an approach.
 

FAQ: Prove that 3n^2 - 1 can't be a square of a integer n

What is the equation being asked to prove?

The equation being asked to prove is 3n^2 - 1 = n^2 + n^2 - 1.

Why is it important to prove this equation?

Proving this equation can help to expand our understanding of number theory and provide insights into the properties of numbers.

Can you provide an example to illustrate the equation?

For example, if n = 3, then 3n^2 - 1 = 3(3)^2 - 1 = 27 - 1 = 26. This is not a perfect square, as there is no integer value for n that can result in a perfect square when substituted into the equation.

What is the mathematical proof for why 3n^2 - 1 can't be a square of an integer n?

First, we can rewrite the equation as 3n^2 - 1 = (n^2 - 1) + 2n^2. Since n^2 - 1 is always an even number, and 2n^2 is always an even number, the sum of these two terms will always be an even number. However, a perfect square must be an odd number multiplied by an odd number. Therefore, 3n^2 - 1 cannot be a perfect square.

Are there any exceptions to this proof?

No, there are no exceptions to this proof. 3n^2 - 1 can never be a perfect square, regardless of the value of n.

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