congruent number

The question of determining whether a given rational number is a congruent number is called the congruent number problem. {{as of|2019}}, this problem has not been brought to a successful resolution. Tunnell's theorem provides an easily testable criterion for determining whether a number is congruent; but his result relies on the Birch and Swinnerton-Dyer conjecture, which is still unproven.

Fermat's right triangle theorem, named after Pierre de Fermat, states that no square number can be a congruent number. However, in the form that every congruum (the difference between consecutive elements in an arithmetic progression of three squares) is non-square, it was already known (without proof) to Fibonacci.{{citation|title=Number Theory and Its History|first=Øystein|last=Ore|author-link=Øystein Ore|publisher=Courier Dover Corporation|year=2012|isbn=978-0-486-13643-1|pages=202–203|url=https://books.google.com/books?id=beC7AQAAQBAJ&pg=PA202}}. Every congruum is a congruent number, and every congruent number is a product of a congruum and the square of a rational number.{{citation|first=Keith|last=Conrad|title=The congruent number problem|journal=Harvard College Mathematical Review|url=http://www.thehcmr.org/issue2_2/congruent_number.pdf|volume=2|issue=2|date=Fall 2008|pages=58–73|url-status=dead|archive-url=https://web.archive.org/web/20130120090003/http://www.thehcmr.org/issue2_2/congruent_number.pdf|archive-date=2013-01-20}}. However, determining whether a number is a congruum is much easier than determining whether it is congruent, because there is a parameterized formula for congrua for which only finitely many parameter values need to be tested.{{citation|title=The Universal Book of Mathematics: From Abracadabra to Zeno's Paradoxes|first=David|last=Darling|publisher=John Wiley & Sons|year=2004|isbn=978-0-471-66700-1|page=77|url=https://books.google.com/books?id=HrOxRdtYYaMC&pg=PA77}}.

n is a congruent number if and only if the system

:$x^2\; -\; n\; y^2\; =\; u^2$, $x^2\; +\; n\; y^2\; =\; v^2$

has a solution where $x,\; y,\; u$, and $v$ are integers.{{cite book |title = Elementary Number Theory |first1 = J. V. |last1 = Uspensky |first2 = M. A. |last2 = Heaslet |author-link1 = J. V. Uspensky |volume = 2 |publisher = McGraw Hill |year = 1939 |page = 419 }}

Given a solution, the three numbers $u^2$, $x^2$, and $v^2$ will be in an arithmetic progression with common difference $n\; y^2$.

Furthermore, if there is one solution (where the right-hand sides are squares), then there are infinitely many: given any solution $(x,\; y)$,

another solution $(x\text{'},\; y\text{'})$ can be computed from{{cite book |title = History of the Theory of Numbers |first = Leonard Eugene |last = Dickson |author-link = Leonard Eugene Dickson |volume = 2 |publisher = Chelsea |year = 1966 |pages = 468–469 }}

:$x\text{'}\; =\; (x\; u)^2\; +\; n\; (y\; v)^2,$

:$y\text{'}\; =\; 2\; x\; y\; u\; v.$

For example, with $n\; =\; 6$, the equations are:

:$x^2\; -\; 6\; y^2\; =\; u^2,$

:$x^2\; +\; 6\; y^2\; =\; v^2.$

One solution is $x\; =\; 5,\; y\; =\; 2$ (so that $u\; =\; 1,\; v\; =\; 7$). Another solution is

:$x\text{'}\; =\; (5\; \backslash cdot\; 1)^2\; +\; 6\; (2\; \backslash cdot\; 7)^2\; =\; 1201,$

:$y\text{'}\; =\; 2\; \backslash cdot\; 5\; \backslash cdot\; 2\; \backslash cdot\; 1\; \backslash cdot\; 7\; =\; 140.$

With this new $x\text{'}$ and $y\text{'}$, the new right-hand sides are still both squares:

:$u\text{'}^2\; =\; 1201^2\; -\; 6\; \backslash cdot\; 140^2\; =\; 1324801\; =\; 1151^2,$

:$v\text{'}^2\; =\; 1201^2\; +\; 6\; \backslash cdot\; 140^2\; =\; 1560001\; =\; 1249^2.$

Using $x\text{'}=1201,\; y\text{'}=140,\; u\text{'},\; v\text{'}$ as above gives

:$u\text{'}\text{'}=1,727,438,169,601$

:$v\text{'}\text{'}=2,405,943,600,001$

Given $x,\; y,\; u$, and $v$, one can obtain $a,\; b$, and $c$ such that

:$a^2\; +\; b^2\; =\; c^2$, and $\backslash frac\{ab\}\{2\}\; =\; n$

from

:$a\; =\; \backslash frac\{v\; -\; u\}\{y\},\; \backslash quad\; b\; =\; \backslash frac\{v\; +\; u\}\{y\},\; \backslash quad\; c\; =\; \backslash frac\{2x\}\{y\}.$

Then $a,\; b$ and $c$ are the legs and hypotenuse of a right triangle with area $n$.

The above values $(x,\; y,\; u,\; v)\; =\; (5,\; 2,\; 1,\; 7)$ produce $(a,\; b,\; c)\; =\; (3,\; 4,\; 5)$. The values $(1201,\; 140,\; 1151,\; 1249)$ give $(a,\; b,\; c)\; =\; (7/10,\; 120/7,\; 1201/70)$. Both of these right triangles have area $n\; =\; 6$.

The question of whether a given number is congruent turns out to be equivalent to the condition that a certain elliptic curve has positive rank. An alternative approach to the idea is presented below (as can essentially also be found in the introduction to Tunnell's paper).

Suppose {{mvar|a}}, {{mvar|b}}, {{mvar|c}} are numbers (not necessarily positive or rational) which satisfy the following two equations:

:$\backslash begin\{align\}$

a^2 + b^2 &= c^2, \\

\tfrac{1}{2}ab &= n.

\end{align}

Then set {{math|x {{=}} n(a + c)/b}} and

{{math|y {{=}} 2n²(a + c)/b²}}.

A calculation shows

:$y^2\; =\; x^3\; -\; n^2\; x$

and {{mvar|y}} is not 0 (if {{math|y {{=}} 0}} then {{math|a {{=}} −c}}, so {{math|b {{=}} 0}}, but {{math|({{frac|1|2}})ab {{=}} n}} is nonzero, a contradiction).

Conversely, if {{mvar|x}} and {{mvar|y}} are numbers which satisfy the above equation and {{mvar|y}} is not 0, set

{{math|a {{=}} (x² − n²)/y}},

{{math|b {{=}} 2nx/y}}, and {{math|c {{=}} (x² + n²)/y}}. A calculation shows these three numbers

satisfy the two equations for {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} above.

These two correspondences between ({{mvar|a}},{{mvar|b}},{{mvar|c}}) and ({{mvar|x}},{{mvar|y}}) are inverses of each other, so we have a one-to-one correspondence between any solution of the two equations in

{{mvar|a}}, {{mvar|b}}, and {{mvar|c}} and any solution of the equation in {{mvar|x}} and {{mvar|y}} with {{mvar|y}} nonzero. In particular, from the formulas in the two correspondences, for rational {{mvar|n}} we see that {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} are rational if and only if the corresponding {{mvar|x}} and {{mvar|y}} are rational, and vice versa.

(We also have that {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} are all positive if and only if {{mvar|x}} and {{mvar|y}} are all positive;

from the equation {{math|y² {{=}} x³ − xn² {{=}} x(x² − n²)}}

we see that if {{mvar|x}} and {{mvar|y}} are positive then {{math|x² − n²}} must be positive, so the formula for {{mvar|a}} above is positive.)

Thus a positive rational number {{mvar|n}} is congruent if and only if the equation

{{math|y² {{=}} x³ − n²x}} has a rational point with {{mvar|y}} not equal to 0.

It can be shown (as an application of Dirichlet's theorem on primes in arithmetic progression)

that the only torsion points on this elliptic curve are those with {{mvar|y}} equal to 0, hence the

existence of a rational point with {{mvar|y}} nonzero is equivalent to saying the elliptic curve has positive rank.

Another approach to solving is to start with integer value of n denoted as N and solve

:$N^2\; =\; ed^2\; +\; e^2$

where

:$\backslash begin\{align\}$

c &= n^2/e + e\\

a &= 2n\\

b &= n^2/e - e

\end{align}

For example, it is known that for a prime number {{mvar|p}}, the following holds:{{citation |author=Paul Monsky |author-link=Paul Monsky |title=Mock Heegner Points and Congruent Numbers |journal=Mathematische Zeitschrift |volume=204 |issue=1 |year=1990 |pages=45–67 |doi=10.1007/BF02570859|s2cid=121911966 }}

• if {{math|p ≡ 3 (mod 8)}}, then {{mvar|p}} is not a congruent number, but 2{{mvar|p}} is a congruent number.
• if {{math|p ≡ 5 (mod 8)}}, then {{mvar|p}} is a congruent number.
• if {{math|p ≡ 7 (mod 8)}}, then {{mvar|p}} and 2{{mvar|p}} are congruent numbers.

It is also known that in each of the congruence classes {{math|5, 6, 7 (mod 8)}}, for any given {{mvar|k}} there are infinitely many square-free congruent numbers with {{mvar|k}} prime factors.{{citation

| last = Tian | first = Ye | author-link = Tian Ye (mathematician)

| arxiv = 1210.8231

| doi = 10.4310/CJM.2014.v2.n1.a4

| issue = 1

| journal = Cambridge Journal of Mathematics

| mr = 3272014

| pages = 117–161

| title = Congruent numbers and Heegner points

| volume = 2

| year = 2014| s2cid = 55390076 }}.

{{Reflist|2}}

• {{citation |last=Alter |first=Ronald |author-link=Ronald Alter |title=The Congruent Number Problem |journal=American Mathematical Monthly |volume=87 |issue=1 |year=1980 |pages=43–45 |doi=10.2307/2320381 |publisher=Mathematical Association of America |jstor=2320381}}
• {{citation |doi=10.1007/BF02837344 |last=Chandrasekar |first=V. |title=The Congruent Number Problem |journal=Resonance |volume=3 |issue=8 |year=1998 |pages=33–45 |s2cid=123495100 |url=http://www.math.rug.nl/~top/Chandrasekar.pdf}}
• {{citation |last=Dickson |first=Leonard Eugene |author-link=Leonard Eugene Dickson |title=History of the Theory of Numbers |volume=II: Diophantine Analysis |date=2005 |series=Dover Books on Mathematics |publisher=Dover Publications |isbn=978-0-486-44233-4 |chapter=Chapter XVI}} – see, for a history of the problem.
• {{citation |last=Guy |first=Richard |author-link=Richard K. Guy |title=Unsolved Problems in Number Theory |series=Problem Books in Mathematics (Book 1) |edition=3rd |date=2004 |publisher=Springer |isbn=978-0-387-20860-2 | zbl=1058.11001}} – Many references are given in it.
• {{citation

| last = Tunnell

| first = Jerrold B.

| author-link= Jerrold B. Tunnell

| title = A classical Diophantine problem and modular forms of weight 3/2

| journal = Inventiones Mathematicae

| volume = 72

| issue = 2

| pages = 323–334

| year = 1983

| doi = 10.1007/BF01389327

| bibcode = 1983InMat..72..323T

| url = http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002099403

| hdl = 10338.dmlcz/137483

| hdl-access = free

}}

• {{MathWorld |urlname=CongruentNumber |title=Congruent Number}}
• A short discussion of the current state of the problem with many references can be found in Alice Silverberg's [http://www.math.uci.edu/~asilverb/bibliography/pcmibook.ps Open Questions in Arithmetic Algebraic Geometry] (Postscript).
• [http://www.aimath.org/news/congruentnumbers/ A Trillion Triangles] - mathematicians have resolved the first one trillion cases (conditional on the Birch and Swinnerton-Dyer conjecture).

{{Classes of natural numbers}}

Category:Arithmetic problems of plane geometry

Category:Elliptic curves

Category:Triangle geometry

Category:Unsolved problems in number theory