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Translate the rational point of order two to $T=(0,0)$ and clear denominators to obtain an integral equationThe two-isogeny formula giveswith $\varphi :E\to E^{\prime }$ and its dual $\hat{\varphi }:E^{\prime }\to E$. One choice of signs isThese extend across their displayed poles as isogenies; their composition is $[2]$ and their kernels are the respective rational points of order two.

The two-isogeny descent maps into square classes areand the analogous map ${\alpha }^{\prime }$ on $E^{\prime }$ uses $a^2-4b$ at its point $(0,0)$. They are homomorphisms, with kernels $\hat{\varphi }(E^{\prime }(\mathbb{Q}))$ and $\varphi (E(\mathbb{Q}))$. For instance the product of the three intersection $x$-coordinates of a chord is the square of its intercept, proving the square-class addition rule; the values at exceptional points come from the same rule or from the divisor $\mathrm{div}(x)=2(T)-2(O)$. This also identifies the maps as the Kummer maps for the two isogenies.

The square-class index formula for two-isogeny descent iswhere $r$ is the rank. To see the torsion correction explicitly, put $T^{\prime }=(0,0)$ on $E^{\prime }$. Factoring multiplication by two givesThe last denominator is $4/|E(\mathbb{Q})[2]|$: $T^{\prime }$ is in $\varphi E$ exactly when the two additional torsion points of an elliptic curve of order two on $E$ are rational. Since $[E:2E]=2^r|E(\mathbb{Q})[2]|$, cancellation gives the formula in both the single and full rational two-torsion cases.

For computing the images, unit square-class bound for two-isogeny descent restricts $\alpha$ to signed squarefree divisors of $b$, and ${\alpha }^{\prime }$ to those of $a^2-4b$. To justify the prime support, negative valuations of a rational $x$-coordinate are even; at a prime not dividing $b$, a positive valuation of $x$ makes $x^2+ax+b$ a unit, so the equation again forces the valuation to be even. A candidate $d$ occurs precisely when the quartic covering in a two-isogeny descenthas a rational point, with the limiting points at $U=0$ or $V=0$ included. Clearing denominators gives integers with $\gcd (U,V)=1$. For $UV\ne 0$, the associated point is $x=d(U/V{)}^2$, $y=dUN/V^3$. Test these finitely many coverings over $\mathbb{R}$ and over relevant ${\mathbb{Q}}_q$ to eliminate impossible classes; surviving locally soluble classes give an upper bound. Finding rational points on them proves membership and often makes the bound exact. Local solubility alone need not prove global solubility, so one must retain the possible Selmer obstruction.

For the illustration takewith $p\equiv 3(\mathrm{mod}8)$. On $E_p$ all possible square classes are $1,-1,p,-p$. They all occur: $O$ supplies $1$ and the three points with $y=0$, at $x=0,p,-p$, supply $-1,p,-p$. Hence $|\alpha (E_p)|=4$.

On $E_p^{\prime }$, a nonzero real point has $X>0$, because $X(X^2+4p^2)=Y^2$. Thus only $d=1,2,p,2p$ need be considered, and $d=1$ occurs already at $O$. The other three classes are excluded as follows.

For $d=p$, the covering is $N^2=p(U^4+4V^4)$. Reducing modulo $p$ forces $p\mid N$ and $U^4+4V^4\equiv 0$. If $V$ were divisible by $p$, then $U$ would be too, contradicting primitivity. Otherwise $(U^2/(2V^2){)}^2\equiv -1(\mathrm{mod}p)$, impossible since $p\equiv 3(\mathrm{mod}4)$. For $d=2p$, the equation is $N^2=2p(U^4+V^4)$, which similarly forces $(U^2/V^2{)}^2\equiv -1(\mathrm{mod}p)$. Thus neither class occurs. These are odd-prime obstructions for the congruent-number isogeny coverings.

For $d=2$, the equation is $N^2=2(U^4+p^2{V}^4)$. If exactly one of $U,V$ is odd, its right side is $2(\mathrm{mod}8)$, not a square. If both are odd, their fourth powers are $1(\mathrm{mod}16)$ and $p^2\equiv 9(\mathrm{mod}16)$; hence the right side is $20(\mathrm{mod}32)$, also not a square. Both even is forbidden by primitivity. This is the modulo-thirty-two obstruction for a congruent-number descent class. Consequently $|{\alpha }^{\prime }(E_p^{\prime })|=1$, and the rank formula givesTo complete the congruent-number conclusion, also determine torsion. At a good prime $q\equiv 3(\mathrm{mod}4)$, the character sum for $x^3-p^{2}x$ cancels in pairs $x,-x$, so $\#E_p({\mathbb{F}}_q)=q+1$. Rational torsion injects into good reduction at every odd prime, by the formal kernel, which is a torsion-free group. If $p\ne 3$, use $q=3,7$ to bound its order by $\gcd (4,8)=4$. If $p=3$, use $q=7,11$, giving $\gcd (8,12)=4$. The three rational points of order two already give four points, so $E_p(\mathbb{Q})\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$. This is rational torsion of a congruent number curve in the present cases.

A congruent number is the positive area of a right triangle with rational side lengths. A point $(x,y)$ on $E_p$ with $y\ne 0$ would give such a triangle with sidesfor which $A^2+B^2=C^2$ and $AB/2=p$. Conversely a rational right triangle of area $p$ gives a point with nonzero $y$, for example $x=p(C+A)/B$, $y=2p^2(C+A)/B^2$. But every rational point on the curve just determined is $O$ or has $y=0$. Every prime $p\equiv 3(\mathrm{mod}8)$ is therefore not a congruent number.

First compute the Mordell-Weil group rank by two-isogeny descent. HereThe two-isogeny formula gives $\varphi :E\to E^{\prime }$. If $u=1+i$, then $u^4=-4$, andis an isomorphism over $K$. Its $x$-coordinate multiplier $u^2$ is a square in $K$.

Use the two-torsion square-class homomorphismFor any prime ideal of the Gaussian integers, if $v(x)>0$, then $x^2+i$ is a unit and $2v(y)=v(x)$; if $v(x)<0$, the term $x^3$ has strictly smallest valuation and $2v(y)=3v(x)$. Hence every valuation of $x$ is even. Since the Gaussian integers form a principal ideal domain, dividing by a square leaves a unit. Their units are $\{1,-1,i,-i\}$, whose square classes are $\{1,i\}$ because $-1=i^2$ and $i$ is not a square in $K$. For the latter assertion, $(a+bi{)}^2=i$ with $a,b\in \mathbb{Q}$ would imply $a^2=b^2$ and $2ab=1$, hence $a^2=1/2$, impossible for rational $a$. ThusBoth classes occur, at $O$ and $T$. This is the unit square-class bound for two-isogeny descent.

Define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly, with ${\alpha }^{\prime }((0,0))=[-4i]=[i]$. Since $\theta$ multiplies nonexceptional $x$-coordinates by a square, and preserves the exceptional classes as well, its image is also $\{1,[i]\}$. The standard kernel identities in two-isogeny descent areHere $\hat{\varphi }$ is the dual isogeny and $\hat{\varphi }\varphi =[2]$. These identities can be checked directly from the formulas: $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$, and conversely a square $x$-coordinate lets the quadratic equation for a preimage be solved using the curve equation. For example, if $x=r^2\ne 0$, the equation for $X$ is $X^2-(4x+2a)X+(a^2-4b)=0$, whose discriminant is $16(x^2+ax+b)=16(y/r{)}^2$; the $Y$-coordinate then follows from the dual formula. The exceptional points satisfy the same completed square-class criterion.

The two-isogeny index formula over a number field keeps track of a small kernel factor:In this case $\ker \hat{\varphi }=\{O,T^{\prime }\}$, where $T^{\prime }=(0,0)$, and $T^{\prime }\notin \varphi E(K)$ because ${\alpha }^{\prime }(T^{\prime })=[i]\ne 1$. Thus $\delta =2$, and the index is $2\cdot 2/2=2$. There is only one nonzero rational 2-torsion point on $E$: the other two would require $x^2=-i$, and $-i$ has the same nonsquare class as $i$. The Mordell-Weil theorem now givesso $r=0$.

It remains to identify all torsion, rather than merely the rank. The elliptic-curve discriminant is $-64i^3$, so the primes $(3)$ and $(2-i)$ have good reduction, with residue characteristics three and five. By the supplied point-count information their reduction groups have orders that are powers of two. The reduction of torsion points on an elliptic curve is injective on prime-to-residue-characteristic torsion. Every odd-primary torsion subgroup therefore injects into a group of two-power order at at least one of these two primes, and must be zero. All torsion is two-primary.

Finally, if a point had order four, its double would be $T$. The elliptic-curve addition formula givesFor $2P=T$ the denominator is nonzero, so $x^2=i$, impossible in $K$. A point of higher two-power order would have a multiple of order four, so it too is excluded. Thus the only torsion points are $O,T$. Together with rank zero,