It is given that $$ -\sqrt{ \frac{8}{3} } \lt x \lt \sqrt { \frac{8}{3} } ~~~~~~~~~~~~\left( x \in \mathbb Q \right)$$
And we want to prove that: $$ 8 - 3x^2 \textrm{can not be a perfect square}$$
I tried to first convert that $\sqrt { \frac{8}{3} }$ into decimals and found that we can write $$ -1.6329 \lt x \lt 1.6329 $$
So, because of this let’s only consider the absolute values of $x$ and let’s take the value of basic step (that is in general when we prove something by induction we do the first step by checking it at $n=1$) as $x=0$, $$ 8 - 3 \cdot 0^2 = 8 $$ $$ \textrm{Since 8 is not a perfect square, therefore we have established the first step of induction}$$
Let’s assume that $$ 8 - 3x^2$$ is not a perfect square for $x=k$, now let’s check it for $x=k+1$ (here comes the problem, this is the inductive step, is it right to do $k+1$?) $$ 8 - 3 ( k+1)^2$$ $$ 8 - 3k^2 -3 - 6k$$ We know that $ 8- 3k^2$ is not a perfect square but how to argue that subratcting $3+6k$ from it wouldn’t change it’s nature. After all, $k$ should lie between $0$ and $1.6329$ and adding $1$ to $k$ seems wrong to me.
Please help me in constructing the proof.
This is not about an inductive proof. From the given you have $0 \leq 8-3x^2 \leq 8$. So there are only finitely many cases for $8-3x^2$ to be a perfect square:
i) $8-3x^2 = 0 \implies x=\pm\dfrac{\sqrt{8}}{\sqrt{3}}$ which is a contradiction to $x$ being rational.
ii) $8-3x^2 = 1 \implies x=\pm\dfrac{\sqrt{7}}{\sqrt{3}}$ which is a contradiction to $x$ being rational.
iii) $8-3x^2 = 4 \implies x=\pm\dfrac{2}{\sqrt{3}}$ which is a contradiction to $x$ being rational.
Thus, $8-3x^2$ can not be perfect square for a given such $x$.
Edit: For the comments on perfect squares can be of the form $\dfrac{k^2}{m^2}$
Let $x=\dfrac{a}{b}$ with $\gcd(a,b)=1$. Then $8-3x^2 = 8 - \dfrac{3a^2}{b^2} = \dfrac{8b^2 - 3a^2}{b^2}$ so need to have
$8b^2 - 3a^2$ is a perfect square integer. That is, $8b^2 - 3a^2 = n^2$ for some $n \in \mathbb{N}$. Now,
$\displaystyle 8b^2 - 3a^2 = n^2 \implies 2b^2 \equiv n^2 \quad (Mod \, \,3) \implies b \equiv 0 \quad (Mod \, \,3)$ and $ n \equiv 0 \quad (Mod \, \,3)$
Last implication is because any square integer is equivalent to either $0$ or $1$ in $(Mod \, \,3)$. So,
$b^2 \equiv 1 \quad (Mod \, \,3)$ can't be true as it implies $n^2 \equiv 2 \quad (Mod \,3)$ Thus, $b=3k$ and $n=3m$.
From this,
$8b^2 - 3a^2 = n^2 \implies 8(3k)^2 - 3a^2 = (3m)^2 \implies 3.8k^2 - a^2 = 3m^2 \implies a \equiv 0 \quad (Mod \, \,3)$
That is, $a=3j$. But we also found $b=3k$. Hence, this is a contradiction to $\gcd(a,b)=1$.
Consequently, $8-3x^2$ can not even be "rational perfect square" for a given such $x$.