Rational points on $y^2=5(x^4+1)$

Question

How would one find $\mathbb{Q}$ -rational points on $y^2=5(x^4+1)$. I do not think there is one but I would also love to see a proof of it. I thought about writing them in terms of fractions and got the following diophatine equation $$X^2=5Y^2+5Z^2$$

and deduced that one of $Y$ and $Z$ must be $1$ mod $5$ and the other is $4$ mod $5$ but I cannot continue. Any help would be appreciated!

Answer 1

Let $(x,y)$ be a rational point on that curve. Then clearing denominators yields an integral solution $(X,Y,Z)$ to $$Y^2Z^2=5X^4+5Z^4,$$ with $\gcd(X,Y,Z)=1$. Then precisely one of $X$, $Y$ and $Z$ is even.

1. If $X$ is even then $Y$ and $Z$ are odd, and so reducing mod $8$ yields $$1\equiv(YZ)^2\equiv5X^4+5Z^4\equiv5\pmod{8},$$ a contradiction.

2. If $Y$ is even then $X$ and $Z$ are odd, so reducing mod $4$ yields $$0\equiv(YZ)^2\equiv5X^4+5Z^4\equiv2\pmod{4},$$ a contradiction.

3. If $Z$ is even then $X$ and $Y$ are odd, so reducing mod $4$ yields $$0\equiv(YZ)^2\equiv5X^4+5Z^4\equiv1\pmod{4},$$ a contradiction.

This shows that there is no rational point on the curve.