Are there any nonzero integers $a$ , $b$ such that $a^2$ = $3b^2$

Question

I know since 3 is prime then nothing divides 3 except 3 and also 3 is a factor for only multiples of 3.

$a^2$ must be a multiple of 3. But I am kinda stuck here.

Answer 1

No. You can prove that the equality

$$a^2=k\cdot b^2$$ only has a solution for integer $a,b$ when $k$ is a perfect square. Probably the easiest way to see that would be to take the square root of both sides, so $$a=\sqrt{k}\cdot b$$ or, equivalently, $$\sqrt{k}=\frac{a}b$$ meaning that this can be solved exactly when $\sqrt{k}$ is rational. To show that $\sqrt{k}$ is rational only when $k$ is a perfect square just notice that, if we write any rational $\frac{p}{q}$ in lowest terms, then its square $\frac{p^2}{q^2}$ is also in lowest terms; thus, if $\left(\frac{p}{q}\right)^2$ is an integer, it means $q^2=1$, since an integer in lowest terms has denominator 1. It therefore must be that $q=1$, meaning that $\frac{p}q$ is an integer. So, if an integer $k$ is not the square of an integer, it cannot be the square of any other rational, and hence its square root is irrational. Obviously, $1^2<3<2^2$ so $3$ is not a perfect square.

Answer 2

If $\frac{a^2}{b^2}$ is an integer, the integer has to be a perfect square, which $3$ is not. Hence, no such $a$ and $b$ exist.

Answer 3

Your equation implies $a=\sqrt 3b$ or $ a=- \sqrt 3b$

There are no integers which satisfy this equation.

A long ago people thought that every number can be expressed in a/b form but no that was proved to be wrong so again number systems where enlarged to include even irrational numbers which gave rise to the very important number system the real number system.