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The change of variables $X=5x$, $Y=25y$ identifies $E_5$ withA full two-isogeny descent, equivalently the standard full 2-descent for a cubic with three rational roots, givesThe local conditions at $2$, $5$, and infinity leave exactly these eight square-class combinations; primes outside $\{2,5\}$ have even valuations and contribute none. Hence the quotient has dimension three. Since the rational 2-torsion has dimension two, the rank is one.

The Lutz–Nagell theorem on the integral model, or reduction at two good primes, excludes odd torsion and torsion of order greater than two. Thereforeand $(-4/5,6/25)$ generates the free part. This is the Mordell-Weil group of the congruent number curve for five.

For $E:y^2=x^3-x^2+25x$, the addition formulas giveIn particular, $[2]Q=T$ and $Q$ has order four, while the nonintegral coordinate of $[2]P=(144/25,-2172/125)$ and part (a) show that $P$ has infinite order.

For two-isogeny descent, writeThe rational point $(0,0)$ is the kernel of a degree-two isogeny $E\to E^{\prime }$. The square-class maps send an affine point with $x\ne 0$ to $[x]\in {\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$, send $O$ to $1$, and send $(0,0)$ to $[b]$ or $[a^2-4b]$ on the two curves. Their images determine the rank through

HereOn $E$, the possible square classes divide $25$; real solubility excludes the negative classes, while $P$ and $Q$ exhibit $1$ and $5$. Thus $\alpha (E(\mathbb{Q}))=\{1,5\}$. On $E^{\prime }$, the pointsexhibit the classes $-11,-1,11$, along with $1$. The remaining candidate classes are divisible by $3$. Their homogeneous spaceshave no primitive solution modulo $9$: reduction modulo $3$ first forces $3\mid UV$, and then the equation is congruent to $3$ or $6$ modulo $9$. HenceThe rank formula gives $2^r=2\cdot 4/4=2$, so $r=1$.

At the good primes $7$ and $19$, direct counts giveThe reduction of torsion points on an elliptic curve injects rational torsion into both groups, so its order divides $\gcd (12,16)=4$. Since $Q$ has order four, the torsion subgroup is $\mathbb{Z}/4\mathbb{Z}$. Thereforeso $t=1$, $d_1=4$, and $r=1$.

For $E:y^2=x^3+a{x}^2+bx$ with rational 2-torsion $(0,0)$, the quotient by that point is the two-isogenous curveThe two-isogeny descent maps a nonexceptional point to the square class of its $x$-coordinate, with $(0,0)$ mapping to $[b]$. The images are finite collections of squarefree divisors of $b$ and $a^2-4b$, determined by testing the associated homogeneous quartics for rational points. If their orders are $2^s$ and $2^{s^{\prime }}$, thenwhich determines the Mordell-Weil rank $r$.

The method requires a rational 2-isogeny, and deciding whether every locally soluble quartic is globally soluble can be difficult. Computing only local conditions gives a 2-isogeny Selmer group and hence an upper bound; a nontrivial Tate-Shafarevich group can make that bound strict. Even after finding the rank, a separate saturation and point search may be needed to find generators.

For this curve, the square-class image in the first two-isogeny descent is contained in $\{1,-1,11,-11\}$. All four classes occur: $Q=(-1,1)$ gives $-1$, $T=(0,0)$ gives $11$, and $Q+T$ gives $-11$. The isogenous curve isIts image is contained in $\{1,-1,5,-5\}$. Negative $x$ cannot occur because $x(x^2-26x+125)<0$ for $x<0$, while $1$ and $5$ are represented by the identity and $(0,0)$. Thus the two image orders are four and two, andgives $r=1$.

The point $P$ has order three because $2P=-P$, and $T$ has order two, so the rational torsion contains a cyclic subgroup of order six. At the good primes $7$ and $13$, direct point counting givesReduction bounds the rational torsion order by their greatest common divisor, namely six, so this is all the torsion. The structure theorem for finitely generated modules over a principal ideal domain now givesThus one may take $d_1=1$, $d_2=6$, and $r=1$.

Foruse two-isogeny descent throughThe square-class maps send a nonexceptional point to the class of its $x$-coordinate. On $E$, possible classes are $\pm 1,\pm 7$; the defining quartics and positivity exclude the negative classes, while $(1,3)$ and $(7,21)$ realize $1$ and $7$. Thus the image has order two. On $E^{\prime }$, the possible classes are $\pm 1,\pm 3$; the classes $1$ and $-3$ occur, while the quartics for $-1$ and $3$ have no primitive solution modulo $16$. Hence this image also has order two. The two-isogeny descent formulatherefore gives $r=0$.

The displayed curve has good reduction at $5$ and $11$, where direct counting givesReduction injects rational torsion of order prime to these characteristics, so its order divides $\gcd (6,18)=6$. We already have the six distinct pointsSince the rank is zero, these are all the rational points and $E(\mathbb{Q})\cong \mathbb{Z}/6\mathbb{Z}$.

For integral $a,b$ with $b(a^2-4b)\ne 0$, every square class in the image of the two-isogeny descent map on $y^2=x(x^2+ax+b)$ has a signed square-free representative supported on primes dividing $b$. At a prime not dividing $b$, a positive valuation of $x$ must be even from the curve equation; a negative valuation is even because the cubic leading term determines $2v(y)=3v(x)$. Thus its image has at most $2^{\nu (b)+1}$ elements. Apply this also to the two-isogenous curve and use the square-class index formula for two-isogeny descent to obtain $r\le \nu (b)+\nu (a^2-4b)$, where $\nu$ counts distinct prime divisors of the nonzero absolute value.

For an odd prime $p$, let $E_p:y^2=x^3-p^{2}x$. Its two-isogeny descent partner is $E_p^{\prime }:y^2=x^3+4p^{2}x$. The first two-torsion square-class homomorphism has image exactly $\{1,-1,p,-p\}$: prime-support bounds allow only these classes, and the four rational two-torsion points realize all of them. Every nonidentity point on $E_p^{\prime }$ has positive $x$, so its square-class image is contained in $\{1,2,p,2p\}$. The two-isogeny rank formula gives $2^{\mathrm{rank}{E}_p}=\#{\alpha }^{\prime }(E_p^{\prime })\le 4$, hence the rank is at most $2$.

For an elliptic curve over a number field $K$, the Mordell-Weil theorem writes $E(K)$ as a finite torsion subgroup plus a free abelian group of finite rank of an abelian group $r$. This integer is the rank of the elliptic curve over $K$. It can increase under field extension. A two-isogeny descent supplies rank bounds from rational square-class images; the canonical height pairing can certify independence of rational points.

Suppose the ring of integers of a number field is a principal ideal domain, $a,b$ are integral, and $b$ is a unit. On $y^2=x(x^2+ax+b)$ every nonzero $x$-coordinate has even valuation at every finite prime: positive valuation gives $2v(y)=v(x)$, and negative valuation gives $2v(y)=3v(x)$. Thus the two-torsion square-class homomorphism takes values among unit square classes. For the Gaussian integers, their unit group modulo squares has two elements, represented by $1,i$. This can reduce a two-isogeny descent to a very small calculation even though the field contains no ordering.