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We prove that the only finite-order rational point is the origin: $\boxed{E(\mathbb{Q}{)}_{\mathrm{tors}}=\{O\}}$. The identity always has order one, so “no elements of finite order” must be understood as no nonidentity torsion points.

The given discriminant is a unit at $5$ and $7$, so the integral Weierstrass equation of an elliptic curve has good reduction at both primes. For an odd prime $p$, completing the square in $y^2+xy=f(x)$ shows that the number of $y$ above $x$ is determined byA nonzero quadratic residue gives two points, zero gives one, and a nonresidue gives none. Direct evaluation givesIncluding the origin gives $\#\tilde{E}({\mathbb{F}}_5)=10$ and $\#\tilde{E}({\mathbb{F}}_7)=13$.

At a good prime $p$, reduction is injective on torsion of order prime to $p$: a point in its kernel belongs to the formal group, where multiplication by that order is bijective by part(i), so a point killed by it must be zero. Suppose a nonidentity rational point has finite order. Taking a suitable multiple gives a point of prime order $\ell$. If $\ell \notin \{5,7\}$, reduction at both primes forces $\ell$ to divide both $10$ and $13$, impossible. If $\ell =5$, reduction at $7$ forces $5\mid 13$, impossible. If $\ell =7$, reduction at $5$ forces $7\mid 10$, also impossible. This exhausts all primes and proves the claim. The torsion exclusion from good reduction counts explicitly handles the two residual characteristics instead of incorrectly assuming prime-to-$p$ injectivity also controls $p$-torsion.