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Do there exist some non-zero rational numbers $x, y$ such that $x \neq y$ 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 $p>3$ ?

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 $p>3$ ?

Do there exist some non-zero rational numbers $x, y$ such that

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

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 $p>3$ ?

Do there exist some non-zero rational numbers $x, y$ such that $x \neq y$ 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 $p>3$ ?

Do there exist some non-zero rational numbers $x, y$ such that

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

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 $p>3$ ?

Do there exist some non-zero rational numbers $x, y$ such that

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

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 $p>3$ ?