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

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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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Look at remainders mod 4. What are the possible values for n^2 mod 4?
 
Thanks for the reply!

While that probably is one way of looking at the problem, we haven't yet reached modulus in our studies.
 
You don't really need to study modulus to think about remainders after division by 4. Nothing else comes to mind as an approach.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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