# Why $p\equiv 3$ mod $8$ is not a congruent number [closed]

A positive integer $$n$$ is called a "Congruent Number" if it represents the area of a right angled triangle with rational sides.

I know there is an elementary proof that all primes $$p\equiv 3$$ mod $$8$$ are not congruent, but I am not getting it. Can anyone give me the proof? It proceeds along the line of the proof that $$1$$ is not a congruent number as given by Fermat.

• Are not congruent to what? Aug 6, 2021 at 6:44
• Do you refer to this? Aug 6, 2021 at 6:50
• I have edited the question Aug 6, 2021 at 6:58
• The proof of $1$ not being congruent is not so elementary.. Aug 6, 2021 at 7:09
• @AlvinLepik It uses descent procedures.. which one has to apply here, but I do not see how.. Aug 7, 2021 at 8:11

The Congruent Number problem is a good toy example in many cases, in particular this question serves for didactic and historical reasons, since "modern" ideas and topics sometimes have to be refreshed.

Here is the answer to the question, the main "hidden" reference is:

Nagell, L’analyse indéterminée de degré supérieur, page 16

with a long list of quartics that do not admit an infinity of rational points. For the convenience of the reader, this answer translates and fills in the few details. It is hidden, because there is no direct reference to the Congruent Number problem.

The references i have also give grant to Genocchi, 1855, for this result, but i have not found the source. For instance Chandrasekar, The Congruent Number Problem, page 42, Box 2, in the Non-CN part of the Box there is a $$p_3$$ mentioned. See also Hürlimann, Bell’s Ternary Quadratic Forms and Tunnel’s Congruent Number Criterion Revisited, Advances in Pure Mathematics, 5, page 276, §4. Notes on Congruent Numbers:

To conclude the present work, some comments on the obtained results might be of interest for future research in this area. In the era before Tunnel (...), some important results were already known. For example, Genocchi proved in 1855 and 1874 that a prime $$p\equiv 3 \pmod 8$$ and (...) were not congruent (see Dickson (...), pp. 465, 467). Later on, Nagell (...) gave a very elementary proof of the fact that a prime $$p\equiv 3 \pmod 8$$ was not congruent.

Dickson refers to the "hidden" relation covered in this answer starting with Nagell's quartic.

Nagell considers quartics of the shape $$ax^4+b=cy^2$$, and gives a list of tuples $$(a,b,c)$$ so that there are only finitely many $$\Bbb Q$$-rational points (explicitly given with $$x$$ among $$\pm1$$) on them. Two items in the list are:

• $$(1,-1,p)$$ with a prime $$p\equiv 3\pmod8$$, the quartic is thus $$x^4-1=py^2$$,
• $$(1,-1,2q)$$ with a prime $$q\equiv 5\pmod8$$, the quartic is thus $$x^4-1=2qy^2$$.

It is this form of the result on congruent numbers accepting an "elementary" descent argument as follows. We consider the first one only. The idea of proof is taken from loc. cit. - notation are the same. In the equations above, $$x,y$$ are rational numbers. Multiplying with denominators, we have to show equivalently that there is no non-trivial solution in $$\Bbb Z$$ for $$\tag{*} x^4-y^4=pz^2\ ,\qquad \text{ with } x,y,z\in\Bbb Z\ ,\ (x,y)=1\ ,\ z\ne 0\ .$$ Taking modulo eight, $$x^4,y^4\in\{0,1\}$$, $$pz^2\in\{0,3,4\}$$, so both sides are $$0$$ modulo eight, so $$z$$ is even, and both $$x\equiv y$$ are odd (being relatively prime), so one of $$x\pm y$$ is exactly divisible by $$2$$, let it be $$x+y$$ (after replacing $$y$$ by $$-y$$), $$x^2+y^2$$ is also exactly divisible by $$2$$, so we are in one of the two situations: \left\{ \begin{aligned} x + y &= 2pu^2\ ,\\ x - y &= 4v^2\ ,\\ x^2 + y^2 &= 2w^2\ , \end{aligned} \right. \qquad\text{ or }\qquad \left\{ \begin{aligned} x + y &= 2u^2\ ,\\ x - y &= 4pv^2\ ,\\ x^2 + y^2 &= 2w^2\ , \end{aligned} \right. where $$p$$ is either dividing $$x+y$$, fixed by being exactly divisible by two (first case), or dividing $$x-y$$. Here, $$p$$ cannot divide $$x^2+y^2$$, else working in the field $$\Bbb F_p=\Bbb Z/p$$ we obtain either that $$(x/y)^2$$ or $$(y/x)^2$$ is well defined and $$=-1$$, but the quadratic reciprocity gives $$\left(\frac{-1}p\right)=(-1)^{(p-1)/2}=-1$$, contradiction. (Or note that the multiplicative group $$\Bbb F_p^\times$$ has $$(p-1)\equiv 2\pmod8$$ elements, so it has no elements of order four.)

The integers $$u,v,w$$ are in either case pairwise relatively prime.

• (1) In the first case, we solve for $$x,y$$, get $$x =pu^2+2v^2$$, $$y=pu^2-2v^2$$, so $$2w^2 =x^2+y^2=(pu^2+2v^2)^2+(pu^2-2v^2)^2=2(p^2u^4+4v^4)\ ,$$ so $$4v^4 =w^2-p^2u^4=(w+pu^2)(w-pu^4)$$, here $$w,p,u$$ are odd, the only common (proper) divisor of the last two factors is two, so we can split only as $$w+pu^2=2v_1^4$$, $$w-pu^2=2v_2^4$$, giving $$pu^2=v_1^4-v_2^4$$, and the contradiction, when taken modulo eight. (Here, $$u^2$$ is an odd square, thus $$u^2\equiv 1\pmod 8$$, $$pu^2\equiv 3\pmod 8$$. But the difference of two fourth powers, each either $$0$$ or $$1$$ $$\pmod8$$ cannot produce the $$3$$.) So the first case cannot be realized.

• (2) In the second case we proceed in the same manner. We solve for $$x,y$$, get $$x =u^2+2pv^2$$, $$y=u^2-2pv^2$$, so $$2w^2 =x^2+y^2=(u^2+2pv^2)^2+(u^2-2pv^2)^2=2(u^4+4p^2v^4)\ ,$$ so $$u^4 =w^2-4p^2v^4=(w+2pv^2)(w-2pv^4)$$, here $$w,u$$ are odd, the last two factors are relatively prime, so we can split only as $$w+2pv^2=u_1^4$$, $$w-2pv^2=u_2^4$$, giving $$p(2v)^2=u_1^4-u_2^4\ ,$$ and this is a new solution of the given equation. Is $$2v$$ smaller in absolute value than $$z$$, the variable in the "same place" from the starting solution? Yes: $$p(2v)^2= 4pv^2 = x-y <(x-y)(x+y)(x^2+y^2)=pz^2\ .$$ By infinite descent we have a contradiction. This concludes the proof.

$$\square$$

Now the answer has to establish the connection to the Congruent Number Problem for the given prime $$p\equiv 3\pmod 8$$. For this, let us start with $$x,y\in \Bbb Q$$ with $$x^4-1 =py^2\ ,$$ and consider the (birational) substitution to the variables $$X,Y,Z$$ with \begin{aligned} X &= \frac{4x^2}{y^2}\ ,\qquad Y = \frac{4x(x^4+1)}{y^3}\ .\\[3mm] &\qquad\text{ Then we have:}\\[2mm] X^3 +4p^2X &=X(X^2 + 4p^2)\\ &=\frac1{y^6}\cdot 4x^2\; (16x^2+4p^2y^4)\\ &=\frac1{y^6}\cdot 4x^2\; 4(4x^2+(py^2)^2)\\ &=\frac1{y^6}\cdot 4x^2\; 4(4x^2+(x^4-1)^2)\\ &=\frac1{y^6}\cdot 4x^2\; 4(x^4+1)^2\\ &=Y^2\ . \end{aligned} This leads to a point on the curve $$E$$ with affine equation $$Y^2=X^3+4p^2X$$.

We want the connection to the elliptic curve $$E$$ given by $$v^2=u^3-p^2u$$, used to detect if $$p$$ is a congruent number or not.

$$E$$ has positive rank, iff the isogenous curve obtained by sending the torsion point $$T=(u_0,v_0)=(0,0)$$ to "zero" has positive rank, and we obtain $$E'$$. To see this, consider the substitution: \begin{aligned} X&=\frac{v^2}{u^2}\ ,\qquad Y=\frac{v(u^2+p^2)}{u^2}\ ,\\[2mm] &\qquad\text{ and to get the relation between X and Y we compute:}\\ Y^2 &=\frac{v^2(u^2+p^2)^2}{u^4}\\ &=\frac{v^2(u^2-p^2)^2}{u^4} +\frac{v^2\cdot 4p^2u^2}{u^4}\\ &=\frac{v^2\;u^2(u^2-p^2)^2}{u^6} + 4p^2\frac{v^2}{u^2}\\ &=\frac{v^6}{u^6}+ 4p^2\frac{v^2}{u^2}\\ &=X^3+4p^2X\ . \end{aligned}

So this would be the "elementary answer".

• The above answer was found and typed as it is, no time for improving. But after knowing it as it is, there is a feeling that the presentation is not optimal "for a book". (Books hide searching for the solution path or computational details, they rearrange details and show a dry quick way to proceed, as if everybody would calculate that way first time.) Some question arise, and as such one may consider them as homework. (My computer, sage, did some homework.) Here are the questions. Do we need to go through the isogenous curve $$Y^2 =X^3 +4p^2X$$ with the given "simple" birational map, or there is a simpler map straightforward to the curve $$v^2 =u^3-p^2u$$? The curve $$v^2=u^3-p^2u$$ may be considered by the purist as not the final step, instead the final (and only needed) step would be to have (birationally) for a solution (over some field) of the equation consider by Nagell, $$x^4-1=py^2$$, the sides of the right triangle with area $$p$$.
• The less elementary, but structural solution is starting with the curve $$E_p$$ over $$\Bbb Q$$ with affine equation $$y^2=x^3-p^2x$$ and goes through a decent $$2$$-descent for it.