Can the product of a square number and 3 still be a square number? This will help me answer the "Is there a perfect cuboid?" question

I have tried it with the first 100 square numbers, this is going to be hard.

\(\sqrt{3(0^2)} = 0\)

That probably doesnt help any ahaha...

I've been thinking about this some more.

\(3a^2 = b^2\)

The question is, "Are there any integer values of a and b that make this equation true?"

Zero works but I want to find something besides zero.

I can rewrite it like this:

\(b = \sqrt{3a^2} \\ b = (\sqrt{3})a\)

b needs to be an integer, so what integer for a can make b an integer?

Well since √3 is irrational, the only possible thing that, when multiplied by √3 comes out with something rational will be √3.

√3 times any integer will just be that integer times √3.

You could multiply by √(27), but all that is doing is multiplying by (√3√3√3). There must be an odd number of √3's in a to cancel out the first √3. And if there's an odd number of √3's in a, then a is irrational too and not an integer.

I think it is impossible with anything besides 0.