Do there exist some non-zero rational numbers $x, y$ such that $x \neq \pm y$ and


$$4y^p = x^2 + 3 \tag{1}$$


for some odd prime $p$?


If one lets $y=a/b$ and $x=c/d$ where $a, b, c, d$ are non-zero integers and $\gcd(a, b)=1=\gcd(c, d)$, you obtain $\frac{a^p}{b^p} = \frac{c^2 + 3d^2}{4d^2}$ thus both $c^2 + 3d^2$ and $2d$ must be integral $p-th$ powers. That is, $d=u^p/2$ and


$$v^p = c^2 + 3d^2 = c^2 + 3(u^p /2)^2 \tag{2}$$


for some coprime integers $u, v$. Now, Euler famously demonstrated that (2) is false for $p=3$. Does his argument also work for $p>3$ ?



On the equation $4y^p= x^2 + 3$
Note that $(y,p,x)=(7,3,37)$ is a solution since $4(7^3) = 37^2 + 3 = 1372$.


You got that


$$\frac{a^p}{b^p} = \frac{c^2 + 3d^2}{4d^2}$$


Since $\gcd(c,d) = 1$, if $d$ is even, then $\gcd(4d^2, c^2 + 3d^2) = 1$, so the rest of your analysis would then be correct.


However, if $d$ is odd (e.g., $d = 1$ in my example solution), then $c$ can't be even (if it was, then the numerator would be odd, with just $2$ factors of $2$ in the denominator, which isn't possible for $p \gt 2$), so $c$ must also be odd. In that case, we instead get that $c^2+3d^2$ is a multiple of $4$, with $\gcd\left(\frac{c^2+3d^2}{4},d^2\right) = 1$. Thus, $\frac{c^2+3d^2}{4}$ and $d^2$ are both integral $p$-th powers so, since $p$ is odd, then $d$ is also a $p$-th power. With $d = u^p$, your $(2)$ would then instead be


$$v^p = \frac{c^2+3d^2}{4} = \left(\frac{c}{2}\right)^2 + 3\left(\frac{u^p}{2}\right)^2$$


