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Altitude $\gamma$, also called elevation angle, is the signed angle from the astronomical horizon to a viewing direction, measured in the vertical plane containing that direction and the zenith. Its range is $-90^{\circ }\le \gamma \le 90^{\circ }$: positive altitude is above the horizontal plane, zero altitude lies on the astronomical horizon, and $90^{\circ }$ points to the zenith. Altitude measures direction, rather than a Euclidean distance or a height in metres. In the local horizontal coordinate system, $\sin \gamma =\mathbf{n}\cdot {\mathbf{e}}_U$; the dot product makes the sign unambiguous even for directions below the astronomical horizon.

A concrete motivating observation is the Exploratorium Teacher Institute's Height Sight activity. Its inclinometer uses a sighting tube fixed to a card and an angular scale, with a string and small weight hanging from the scale's pivot. A partner reads where the freely hanging string meets the scale. Sighting a level target should give a readout near zero; sighting a taller target gives an angular height. The activity asks the observer to move closer and farther: the closer position gives the larger angle. The interpretation is that the readout measures the tube's tilt relative to the gravity reference, while converting that angle to a physical height also requires distance and eye height. The activity also extends the inclinometer to sky observations. ^[ref].

The following calibration and sky sequence is a proposed extension, with no claimed observing dataset. First establish two reference marks at the same height with a water level, rather than assuming that a sloping floor or distant ridge is horizontal. Keep the inclinometer in a vertical plane, sight between the marks, wait for the weight to settle, and read the zero several times. Next sight a mark whose independently measured rise $q$ above the tube and horizontal separation $d$ are known; a right triangle predicts ${\gamma }_{\mathrm{ref}}=\arctan (q/d)$. A synthetic $q=3\mathrm{m}$, $d=5\mathrm{m}$ gives ${\gamma }_{\mathrm{ref}}=30.964^{\circ }$. Checking both zero and this nonzero angle can reveal a wrongly attached tube or a scale read from the wrong end. Use level separation, not distance along inclined ground.

If the scale's indicated altitude is $m={\gamma }_{\mathrm{app}}+b$, estimate its additive zero $b$ from the level sight, then obtain apparent altitude ${\gamma }_{\mathrm{app}}=m-b$. For synthetic readings $b=1.0^{\circ }$ and $m=32.0^{\circ }$, the corrected altitude is $31.0^{\circ }$, close to the $30.964^{\circ }$ calibration prediction. Declaring illustrative readout bounds of $0.5^{\circ }$ for each reading gives a conservative combined bound of $1.0^{\circ }$ on $m-b$; these bounds are assumptions, not measured performance. Repeatability tests the variation between readings, whereas a persistent zero offset remains a systematic measurement error. Include uncertainty in the reference rise and separation before judging agreement.

At a fixed site, sight a recognizable star, record the observation time, corrected altitude and north-through-east azimuth, and repeat over a short interval. Record the terrain boundary at the same azimuth separately from the gravity-defined zero. The fixed-direction model in horizontal coordinate system predicts changing angular readouts as the Earth rotates; this is a prospective test, rather than a fabricated measurement curve. Keep the tube aligned, the string free of contact and shielded from wind, the eye at the same reference position, and repeat the level check afterward. Timestamp each sight so genuine sky motion is not mistaken for poor repeatability. An optical telescope with an angular readout can replace the sighting tube, but its mounting still needs an independently established level and angular zero.

The corrected instrument reads apparent altitude, including atmospheric refraction. Under ordinary conditions atmospheric refraction raises a low direction, so if $R$ is the upward angular correction, ${\gamma }_{\mathrm{app}}={\gamma }_{\mathrm{geom}}+R$. Subtracting the instrument zero does not subtract $R$. Record site height, temperature and pressure if applying a refraction model: these are explicit observer inputs in the US Naval Observatory's NOVAS software. A coarse inclinometer may be unable to resolve the correction, and a local weather reading alone does not determine every atmospheric layer. ^[ref].

The circumcircle of a nondegenerate triangle is the unique circle through its three vertices. Its center is the circumcenter, found by intersecting two perpendicular bisectors; equality of distances to all three vertices proves existence, and subtracting the squared-distance equations proves uniqueness. For a right triangle, the midpoint of its hypotenuse is equidistant from all vertices, so its circumradius is half the hypotenuse length.

A conical pendulum is a pendulum whose bob executes uniform horizontal circular motion beneath a fixed pivot, with its string at a constant angle $\theta$ to the downward vertical. The string sweeps a right circular conical surface; the bob follows its circular rim, without any physical cone supporting it. The ideal model has a point mass $m>0$, a taut, massless, inextensible string of length $\ell >0$, uniform gravitational acceleration of magnitude $g>0$, and negligible drag force and pivot friction. A nonzero circular motion requires $0<\theta <\pi /2$. This geometry and the corresponding forces are described in Richard Fitzpatrick, The conical pendulum, equations 262–267.

The right triangle formed by the pivot, the orbit centre, and the bob gives $r=\ell \sin \theta$ and $h=\ell \cos \theta$. Here $r$ is the orbit radius and $h$ is the vertical drop from the pivot to the bob's horizontal plane, not the bob's height above a table. Write $T$ for tension, $P$ for the oscillation period of one revolution, and $\omega =2\pi /P$ for the angular speed; then the bob's speed is $v=\omega r$. A side view compresses the horizontal circle to a line segment; the top view reveals the circle and its tangent velocity.

The two actual forces on the bob are tension toward the pivot and gravitational weight vertically downward. The vertical acceleration is zero, while the horizontal centripetal acceleration is $v^2/r={\omega }^{2}r$. Thus Newton's second law givesThe inward net force is the horizontal component of tension, rather than an extra interaction to add to these two forces. Constant speed still requires acceleration because the velocity changes direction. ^[ref]

Dividing the two component equations and using $r=\ell \sin \theta$ gives the useful collection of predictionsIf $\varphi$ is the azimuthal angle, then $\dot{\varphi }=\pm \omega$; either direction of rotation satisfies ${\dot{\varphi }}^2\cos \theta =g/\ell$. The mass cancels from the angular speed and oscillation period, but not from the tension. The motion is a relative equilibrium: in a rotating frame following the bob at angular speed $\omega$, its position is fixed. This is a concrete way to see how rotational symmetry can turn an apparently moving solution into a stationary one.

For a real angle $\theta$, $\cos \theta$ is the horizontal coordinate on the unit circle reached by a counterclockwise rotation through $\theta$ radians. For an acute angle in a right triangle, it is the adjacent side divided by the hypotenuse. Basic values includeAt $\theta =2\pi /3$, the horizontal coordinate is $-1/2$ even though the vertical sine coordinate remains positive. A positive angle need not produce a positive cosine.

Reflection across the horizontal axis leaves this coordinate unchanged, giving $\cos (-\theta )=\cos \theta$: cosine is an even function. A full turn gives $\cos (\theta +2\pi )=\cos \theta$. For real arguments, its range is $[-1,1]$ and its zeros of a function occur at $\theta =\pi /2+k\pi$. A quarter-turn shifts between the two coordinate traces, with $\cos \theta =\sin (\theta +\pi /2)$.

Electrical impedance is the ratio of terminal electric potential difference and electric current phasors at one specified positive angular frequency $\omega$. For an electrical circuit modeled as a linear time-invariant system, use the sinusoidal steady state after any decaying transient has settled. With the passive sign convention, positive electric current enters the terminal assigned positive electric potential difference. Define peak complex amplitudes byThe physical $v(t)$ and $i(t)$ are real; the complex numbers encode their amplitudes and wave phases. The unit of electrical impedance is the ohm, $1\Omega =1\mathrm{V}/\mathrm{A}$. Both phasors may instead use root mean square magnitudes, with the same ratio $Z$. A nonlinear diode can generate additional frequencies; a single $Z(\omega )$ then does not describe the whole response.

Write the electrical impedance in Cartesian and polar forms asHere $R$ is the resistive real part, $X$ the signed electrical reactance, and $\varphi$ the voltage-minus-current phase difference, defined modulo a full turn. The Pythagorean theorem gives the modulus from a right triangle in the complex plane. The complex argument gives the rotation from the current phasor to the voltage phasor. Thus $V_{\mathrm{rms}}=|Z|I_{\mathrm{rms}}$, while negative $\varphi$ means electric current leads terminal electric potential difference. For $R>0$, $\varphi =\arctan (X/R)$ lies between $-\pi /2$ and $\pi /2$; outside that case, use the quadrant of the complex number rather than a bare inverse tangent. The geometric magnitude and phase difference are the two pieces of the AC relation.^[ref]

Multiplication by $Z$ acts on phasors, not on instantaneous physical signals. For example, when $i(t)=I_0\cos \omega t$,The second term is in phase quadrature with electric current; it cannot be represented by pointwise multiplication $v(t)=Zi(t)$ when $X\ne 0$. This is the useful compression: one complex multiplication replaces separate amplitude and phase difference calculations, while the real part restores the physical signal.

An ideal resistor, inductor and capacitor have $Z_R=R$, $Z_L=\mathrm{i}\omega L$ and $Z_C=1/(\mathrm{i}\omega C)$. Their electrical impedances add in a series RLC circuit, giving $Z=R+\mathrm{i}(\omega L-1/(\omega C))$. The current amplitude is $V_0/|Z|$, so cancellation of the imaginary parts explains the resonant series RLC response. The component-law derivation and opposite complex exponential convention are in reactance phase convention.

Euclidean geometry describes flat Euclidean space using lengths, angles, and rigid motions. A coordinate realization uses points $x,y\in {\mathbb{R}}^n$, Euclidean distanceand angles determined by the dot product. The coordinate axes are a convenience: translations and rotations change coordinates while preserving these measurements. This makes Euclidean geometry useful for comparing figures and converting measurements along different directions into a single distance.

In the Euclidean plane, through a point outside a line there is exactly one parallel line. This is a standard equivalent formulation of the parallel postulate. Euclid's fifth postulate in the Elements instead describes when two lines meet using the interior angles made by a crossing line.^[ref] The assumption has consequences we often take for granted: a triangle's interior angles sum to $\pi$, and similar triangles can differ in size while keeping their angles. Comparing with spherical geometry reveals that these are properties of a particular geometry.

As a computed example, the diagonal from $(0,0)$ to $(4,3)$ in a rectangle has length $\sqrt{4^2+3^2}=5$. The Pythagorean theorem explains why perpendicular contributions add as squares. If the directions cease to meet at a right angle, the law of cosines supplies the missing angle-dependent term. The beauty is that a single distance formula unites geometric constructions, coordinate calculations, and orthogonality in linear algebra.

Make the assumptions visible: in a Cartesian coordinate system with orthogonal axes and the same length unit on each axis, take $A=(0,0)$, $B=(4,0)$ and $C=(0,3)$. The vectors $B-A=(4,0)$ and $C-A=(0,3)$ have dot product zero, establishing a right angle at $A$ before using the Pythagorean theorem. A sketch that merely looks like a right triangle supplies no such guarantee. Likewise, an image stretched horizontally and vertically by different amounts can distort both lengths and angles; a displayed shape and the assumptions defining it must be kept separate.

For a less origin-dependent measurement, take $P=(1,2)$ and $Q=(5,5)$. Their displacement vector is $Q-P=(4,3)$, so their Euclidean distance is again $5$. Its Euclidean norm records the direct separation, whereas travelling first horizontally and then vertically covers length $4+3=7$. This connects measuring a diagonal to locating two positions from a Cartesian coordinate system. A translation changes where the positions are recorded without changing their separation; a rotation changes the component values without changing the Euclidean norm.

There is no positive integer right triangle whose area is a square. For a hypothetical primitive triangle of minimal hypotenuse, its primitive Pythagorean parametrization makes $m,n,m-n,m+n$ pairwise coprime squares. Write $m=r^2$, $n=s^2$, $m+n=u^2$, $m-n=v^2$. Modulo eight forces $m$ odd and $n$ even. The new integer legs $(u+v)/2,(u-v)/2$ have hypotenuse $r$ and area $(s/2{)}^2$, a smaller positive hypotenuse, contradicting infinite descent. Clearing denominators extends the conclusion to rational right triangles of area one.

The field of view is the region of object directions sampled by an optical instrument in one pointing. State whether it is horizontal, vertical or diagonal, and whether it is a full angle or a half-angle. An object-plane width in millimetres also describes a field of view, but requires a specified object distance. Neither that width nor the angular field of view is the aperture diameter or the diffraction-limited resolution.

For a centred flat rectangular photodetector of width $w$ in the focal plane, with positive effective focal length $f$, infinity focus and ideal rectilinear imaging without optical distortion, the full horizontal field of view isThe model maps an object direction $\theta$ to image height of magnitude $|y|=f|\tan \theta |$; set $|y|=w/2$ and use the inverse tangent. This retains exact right triangle geometry within the stipulated imaging model; a paraxial focal length alone does not guarantee that a real wide-field optical lens obeys it. With detector height $h$, replace $w$ by $h$ for the vertical field of view, and by $\sqrt{w^2+h^2}$ for the diagonal field of view. The small-angle approximation gives ${\Theta }_x\simeq w/f$ in radians. ^[ref].

For an original worked example, a $36\mathrm{mm}$ detector width gives ${\Theta }_x\approx 39.60^{\circ }$ at $f=50\mathrm{mm}$, and $20.41^{\circ }$ at $f=100\mathrm{mm}$. Keeping $D=25\mathrm{mm}$ makes these f/2 and f/4 respectively: the field of view changes because $f$ changes. By contrast, closing $D$ from $25$ to $12.5\mathrm{mm}$ at $f=100\mathrm{mm}$ changes f/4 to f/8 while retaining $20.41^{\circ }$ in this model. The clean separation is useful: entrance pupil area controls collection, focal ratio controls the paraxial cone, and $w/f$ controls angular image coverage. Edge clipping, optical distortion, field curvature and optical aberrations can restrict the practically usable field of view.

A proposed angular readout on an optical bench illuminates a small asymmetric target at the optical focus of a collimator, providing a bundle with a known direction. Rotate the optical camera about the centre of its entrance pupil, reading a graduated rotation stage, and record the target image position on its fixed photodetector. Centre the target to establish zero; turn left and right until its centre reaches each detector edge. The difference between stage readings measures the full horizontal field of view. Repeat with a smaller aperture stop: unchanged edge directions but reduced signal support the distinction between collection and angular coverage. Keep the same active detector width, infinity focus, illumination, integration time and gain; approach each reading from the same direction to check mechanical backlash. A finite target blurs the edge criterion, detector thresholds can imitate lost coverage, and a wide bundle must illuminate the whole entrance pupil at every tested angle. These checks distinguish a geometrical boundary from insufficient signal. The analytical diagram is the prediction for this protocol, not a record of its execution.

The hypotenuse is the side opposite the right angle in a right triangle. Its length $c$ is determined by the Pythagorean theorem.

The incircle of a triangle is the circle inside the triangle tangent to all three sides. Its center is the intersection of two internal angle bisectors; its radius is $r=\Delta /s$, where $\Delta$ is the area and $s$ the semiperimeter. To prove it is the largest contained circle, let any candidate have center $P$, radius $\rho$ and distances $d_i\ge \rho$ from the three side lines. Splitting the triangle into three triangles with vertex $P$ gives $\Delta =\frac{1}{2}{\sum }_i{\ell }_i{d}_i\ge s\rho$. The tangent circle attains equality, proving maximality. For a right triangle with legs $a,b$ and hypotenuse $c$, $r=(a+b-c)/2$, since $r=ab/(a+b+c)$ and $a^2+b^2=c^2$.

For $D>0$, a nontorsion rational point on the congruent number elliptic curve $y^2=x^3-D^{2}x$ gives infinitely many positive rational right triangles of area $D$. Apply the formulas $a=|(x^2-D^2)/y|$, $b=|2Dx/y|$, $c=|(x^2+D^2)/y|$ to its multiples. The map is finite-to-one: if $r=c/b$ is fixed, then $|x|$ satisfies $X^2-2DrX+D^2=0$, leaving only finitely many coordinate signs. Thus infinitely many points give infinitely many distinct triangles.

The Mach angle of a supersonic flow with Mach number $M>1$ is $\mu =\arcsin (M^{-1})$. Infinitesimal pressure disturbances lie on the corresponding Mach cone, and a weak oblique shock approaches this angle as its strength tends to zero.

The Mach angle is the half-angle between a Mach cone generator and its backward motion axis for a steady point source in a uniform three-dimensional medium with isotropy and constant sound speed. It obeysThe relation concerns weak acoustic waves in linear acoustics. It is not the angle of a finite oblique shock produced by finite turning of the fluid flow; that angle also depends on the deflection and the gas properties.^[ref]

For a short proof, choose an old emission in the moving-point acoustic wavefronts construction and set $s=-\tau >0$. Its circle has center $(-Us,0)$ and radius $cs$. A tangent line from the present source at the origin meets the circle at a right angle to its radius. The resulting right triangle has hypotenuse $Us$ and opposite side $cs$, giving the displayed sine ratio. The emission age cancels: every wavefront has the same tangent directions, so a continuum of expanding spheres produces one simple Mach cone.

In an original example, stipulate uniform gas with sound speed $c=340\mathrm{m}{\mathrm{s}}^{-1}$ and a source moving through it at $U=680\mathrm{m}{\mathrm{s}}^{-1}$. Then $M=2$ and $\mu =\arcsin (1/2)=30^{\circ }$. An emission one second earlier is now a sphere centered $680\mathrm{m}$ behind the source with radius $340\mathrm{m}$. The selected sound speed defines this example; it is not a constant for air at every temperature. The Mach angle decreases as $M$ increases: its derivative is $-1/(M\sqrt{M^2-1})$ in radians, and its limits are $90^{\circ }$ as $M\downarrow 1$ and $0$ as $M\to \infty$.

For a prime $p\equiv 3(\mathrm{mod}8)$, apply two-isogeny descent to the congruent number elliptic curve $y^2=x^3-p^{2}x$ and its companion $y^2=x^3+4p^{2}x$. The first square-class image is exactly $\{1,-1,p,-p\}$, already supplied by 2-torsion. The second has only positive candidates $1,2,p,2p$. The quartic covering in a two-isogeny descent equations for $p$ and $2p$ are impossible modulo $p$, because $-1$ is not a quadratic residue. For $2$ the equation is $W^2=2U^4+2p^2{V}^4$. Coprimality forces $U,V$ odd and $W=2Z$; modulo sixteen this gives $2Z^2=1+9=10$, which is impossible. Thus the second image is trivial and the square-class index formula for two-isogeny descent gives rank zero. The same prime-reduction argument bounds its rational torsion by four, and it already has four rational 2-torsion points, so it has no rational point corresponding to a right triangle of area $p$.

The parsec is a distance unit defined by the exact conventionThe astronomical unit is exactly $149597870700\mathrm{m}$; one arcsecond is exactly $\pi /648000$ radians. The IAU resolution explicitly records the parsec convention. A megaparsec is one million parsecs. ^[ref]. ^[ref].

The familiar description "the distance giving one arcsecond of stellar parallax" uses the small-angle approximation. Exact right triangle geometry instead gives $\mathrm{au}/\tan (\pi /648000)$, slightly shorter than the conventional parsec. The relative difference is approximately $(\pi /648000{)}^2/3\simeq 7.83\times 10^{-12}$. This is negligible for the examples here, but distinguishing an exact unit convention from an approximate geometric interpretation prevents a misleading equality.

Suppose an integer right triangle with square area exists, and choose one with the least positive hypotenuse $c$. It is primitive. Indeed, dividing all sides by their common divisor $d$ gives an integer right triangle whose area is the square of the rational number $s/d$, where $s^2$ was the original area. Its area is an integer, since one of its legs is even. A rational number whose square is an integer is itself an integer, as follows by writing it in lowest terms. Thus division by $d>1$ would give another square-area triangle with smaller hypotenuse.

By the primitive Pythagorean parametrization, after ordering the legs the triangle has sides $n^2-m^2,2nm,n^2+m^2$, with coprime $n>m>0$ of opposite parity. Its area isThese four positive factors are pairwise coprime: the pair $n-m,n+m$ has gcd dividing $2m$ and is odd, and the other pairs have gcd one by $\gcd (n,m)=1$. Since the product is a square, each factor is a square. WriteBoth $w,z$ are odd. Odd squares are $1(\mathrm{mod}8)$, so $2m=w^2-z^2$ is divisible by eight; hence $m$ is divisible by four and $v$ is even. Define positive integersThey satisfyThus $(r,s,u)$ is another positive integer right triangle with square area, but its hypotenuse $u=\sqrt{n}$ is strictly smaller than $n^2+m^2=c$. This contradicts minimality. No integer right triangle has a nonzero square area. This is the Fermat right triangle theorem, with the actual infinite descent construction exhibited.

Put $q=|z{|}^2$ and retain the domain restriction $z\ne 0$. Then $z+1/\bar{z}=(q+1)/\bar{z}$, so the continued fraction simplifies toThe other denominators are nonzero automatically because $q+1,q+2>0$. Writing $z=x+iy$, the real-part condition becomesMultiplying by $q(q+2)$ and rearranging gives $q+4x+2=0$. Thus the first geometric locus in the complex plane is the circleIts center is $-2$ and radius is $\sqrt{2}$; it does not contain the excluded point $z=0$ on its circumference.

The principal complex logarithm has imaginary part equal to the complex argument, so its equation gives the open rayIf one interprets the logarithm as multivalued, requiring a value with imaginary part $\pi /3$ gives the same ray. Expanding $z^2$ in the other equation gives $2ixy=2ix$, hence $x(y-1)=0$. Its solution is the union of the entire imaginary axis and the horizontal line $y=1$.

These latter loci bound a right triangle with verticesThe bounded open region is $0<x<1/\sqrt{3}$, $\sqrt{3}x<y<1$. Its closure is the closed triangle. Although the logarithm is undefined at $O$, the imaginary-axis locus includes that vertex. The earlier circle is a separate locus and is not a side of this triangle.

The right angle is at $P$, so the circumcenter is the midpoint of hypotenuse $OQ$:Thus its circumcircle has the complex equationTo find the incircle, let its radius be $r$. Tangency to $x=0$ and $y=1$ puts its center at $z_i=r+i(1-r)$. The distance from that point to $y-\sqrt{3}x=0$ is $(1-r-\sqrt{3}r)/2$. Requiring this distance to be $r$ givesThereforeThis tangent circle is indeed the largest contained circle. For any contained disk of radius $\rho$, its center has distances $d_j\ge \rho$ from the three side lines. Splitting the triangle into triangles with that center as vertex gives area $\Delta =\frac{1}{2}{\sum }_j{\ell }_j{d}_j\ge s\rho$, where $s$ is its semiperimeter. Here $\Delta =1/(2\sqrt{3})$ and $s=(1+\sqrt{3})/2$, so $\rho \le \Delta /s=r$. Our circle attains this upper bound, proving maximality rather than just tangency.

A congruent number is a positive integer which occurs as the area of a right triangle with positive rational side lengths. Scaling a triangle by a nonzero rational factor multiplies its area by a rational square, so the essential parameter is its positive square class; integer parameters can be reduced to square-free integers. For example $6$ is congruent by the triangle $(3,4,5)$, and $5$ is congruent by $(3/2,20/3,41/6)$.

The connection with elliptic curves comes from the congruent number elliptic curve in the modelFrom a rational point with $y\ne 0$, defineDirect squaring gives $a^2+b^2=c^2$, and its area isConversely, from a positive rational triangle $(a,b,c)$ of area $N$, putSubstitution using $c^2=a^2+b^2$ and $ab=2N$ gives $y^2=x^3-N^{2}x$ and $y\ne 0$. ThusThe standard rational torsion of a congruent number curve consists of $O,(0,0),(N,0),(-N,0)$. One way to see the torsion restriction is to use good primes $p\equiv 3(\mathrm{mod}4)$. The character sum cancels at $x$ and $-x$, giving $\#E_N({\mathbb{F}}_p)=p+1$. For any odd prime $\ell$, the Dirichlet theorem on primes in arithmetic progressions supplies good primes with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$, so the reduction of torsion points on an elliptic curve excludes $\ell$-torsion. Good primes $p\equiv 3(\mathrm{mod}8)$ similarly bound the two-primary torsion by four, and the four displayed points give equality. Consequently the Mordell-Weil theorem turns the criterion into positive rank of $E_N(\mathbb{Q})$. Arithmetic two-descent on an elliptic curve and local obstructions are therefore tools for the congruent number problem, and a nontorsion point gives infinitely many rational right triangles from a nontorsion point by taking multiples.

We now give elementary proofs for $1$ and $2$, including the descent steps. Recall the primitive Pythagorean parametrizationwhere $m>n>0$ are coprime and of opposite parity. To derive it, in a primitive Pythagorean triple put the even leg at $A$. The relatively prime integers $(C+B)/2,(C-B)/2$ have product $(A/2{)}^2$, so each is a square, say $m^2,n^2$. This gives the formulas and the parity condition. The area is $mn(m-n)(m+n)$, and its four factors are pairwise coprime: their common divisors reduce to divisors of $m,n$ or of two, while $m\pm n$ are odd.

Area one is impossible. Clearing denominators of a hypothetical rational triangle of area one gives an integer triangle with square area. Dividing out its common factor leaves a primitive integer triangle whose area is still a rational square; an integer which is a rational square is an integer square. Choose such a triangle with smallest hypotenuse. Pairwise coprimality in the area product forcesThe positive integers $u,v$ are odd and coprime. If $m$ were even, then $n$ would be odd and $u^2-v^2=2s^2$ would be $2$ modulo eight, whereas two odd squares have difference zero modulo eight. Hence $m$ is odd, $n$ is even and $s$ is even.

Now set $a=(u+v)/2$ and $b=(u-v)/2$. These are positive coprime integers andThus $(a,b,r)$ is another integer right triangle with square area and hypotenuse $r$, strictly smaller than $m^2+n^2$. This contradicts the minimal choice and proves the Fermat right triangle theorem. In particular $1$ is not a congruent number.

For area two we first establish the auxiliary descent fourth-power sum cannot be a square. Suppose $x^4+y^4=z^2$ with positive integers, and choose a primitive solution with smallest $z$. Two odd fourth powers sum to $2$ modulo sixteen, so after exchange $x$ is odd and $y$ is even. Parametrize the primitive triangle $(x^2,y^2,z)$:The case $m$ even, $n$ odd makes $x^2\equiv 3(\mathrm{mod}4)$, so $m$ is odd and $n$ even. Coprimality then gives $m=u^2$ and $n=2v^2$. ConsequentlyThis second triangle is primitive, since $x$ is odd and $\gcd (x,v)=1$. Parametrize it as $x=r^2-s^2$, $2v^2=2rs$, $u^2=r^2+s^2$, with $r,s$ coprime. The equality $v^2=rs$ makes $r=a^2$, $s=b^2$. Thusa new positive solution with smaller hypotenuse $u<z=u^4+4v^4$. This contradicts infinite descent, proving the auxiliary assertion.

Area two is also impossible. Clearing denominators and removing a common factor now gives a primitive integer triangle with area twice an integer square. Indeed, if an integer is $2(a/b{)}^2$ with $\gcd (a,b)=1$, then $b^2$ divides two, forcing $b=1$. Its area product would satisfyThe odd factors must be squares and the unique even factor twice a square. As in the preceding modulo-eight argument, $m$ even would force $m+n=c^2$, $m-n=d^2$ and $n=b^2$ odd, with $c^2-d^2=2b^2$, impossible modulo eight. ThusThe integers $u=(c+d)/2$ and $v=(c-d)/2$ are positive and coprime, and $uv=(c^2-d^2)/4=b^2$. Hence $u=r^2$, $v=s^2$. Butcontradicting the auxiliary descent. This proves two is not a congruent number, and concludesThe two obstructions are global arithmetic descents. The elliptic-curve correspondence organizes the wider problem, but by itself is not a substitute for proving these particular nonexistence assertions.

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.