Congruent number

In mathematics, a congruent number is a positive integer that is the area of a right triangle with three rational number sides.[1] A more general definition includes all positive rational numbers with this property.[2]

Triangle with the area 6, a congruent number.

The sequence of (integer) congruent numbers starts with

5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34, 37, 38, 39, 41, 45, 46, 47, 52, 53, 54, 55, 56, 60, 61, 62, 63, 65, 69, 70, 71, 77, 78, 79, 80, 84, 85, 86, 87, 88, 92, 93, 94, 95, 96, 101, 102, 103, 109, 110, 111, 112, 116, 117, 118, 119, 120, ... (sequence A003273 in the OEIS)

Congruent number table: n ≤ 120

Congruent number table: n ≤ 120
—: non-Congruent number
C: square-free Congruent number
S: Congruent number with square factor

n	1	2	3	4	5	6	7	8
	—	—	—	—	C	C	C	—
n	9	10	11	12	13	14	15	16
	—	—	—	—	C	C	C	—
n	17	18	19	20	21	22	23	24
	—	—	—	S	C	C	C	S
n	25	26	27	28	29	30	31	32
	—	—	—	S	C	C	C	—
n	33	34	35	36	37	38	39	40
	—	C	—	—	C	C	C	—
n	41	42	43	44	45	46	47	48
	C	—	—	—	S	C	C	—
n	49	50	51	52	53	54	55	56
	—	—	—	S	C	S	C	S
n	57	58	59	60	61	62	63	64
	—	—	—	S	C	C	S	—
n	65	66	67	68	69	70	71	72
	C	—	—	—	C	C	C	—
n	73	74	75	76	77	78	79	80
	—	—	—	—	C	C	C	S
n	81	82	83	84	85	86	87	88
	—	—	—	S	C	C	C	S
n	89	90	91	92	93	94	95	96
	—	—	—	S	C	C	C	S
n	97	98	99	100	101	102	103	104
	—	—	—	—	C	C	C	—
n	105	106	107	108	109	110	111	112
	—	—	—	—	C	C	C	S
n	113	114	115	116	117	118	119	120
	—	—	—	S	S	C	C	S

For example, 5 is a congruent number because it is the area of a (20/3, 3/2, 41/6) triangle. Similarly, 6 is a congruent number because it is the area of a (3,4,5) triangle. 3 and 4 are not congruent numbers.

If q is a congruent number then s²q is also a congruent number for any natural number s (just by multiplying each side of the triangle by s), and vice versa. This leads to the observation that whether a nonzero rational number q is a congruent number depends only on its residue in the group

${\displaystyle \mathbb {Q} ^{*}/\mathbb {Q} ^{*2}}$.

Every residue class in this group contains exactly one square-free integer, and it is common, therefore, only to consider square-free positive integers, when speaking about congruent numbers.

Congruent number problem

The question of determining whether a given rational number is a congruent number is called the congruent number problem. This problem has not (as of 2019) 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.[3] Every congruum is a congruent number, and every congruent number is a product of a congruum and the square of a rational number.[4] 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.[5]

Solutions

n is a congruent number if and only if

${\displaystyle x^{2}+nt^{2}=y^{2}}$, ${\displaystyle x^{2}-nt^{2}=z^{2}}$

has solutions (if so, then this equation has infinitely many solutions, like Pell's equation).

Given the solutions {x, y, z, t}, one can obtain the {a, b, c} such that

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

from

${\displaystyle a={\frac {y-z}{t}}}$, ${\displaystyle b={\frac {y+z}{t}}}$, ${\displaystyle c={\frac {2x}{t}}.}$

Relation to elliptic curves

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.[2] An alternative approach to the idea is presented below (as can essentially also be found in the introduction to Tunnell's paper).

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

${\displaystyle {\begin{matrix}a^{2}+b^{2}&=&c^{2},\\{\tfrac {1}{2}}ab&=&n.\end{matrix}}}$

Then set x = n(a+c)/b and y = 2n²(a+c)/b². A calculation shows

${\displaystyle y^{2}=x^{3}-n^{2}x}$

and y is not 0 (if y = 0 then a = -c, so b = 0, but (​¹⁄₂)ab = n is nonzero, a contradiction).

Conversely, if x and y are numbers which satisfy the above equation and y is not 0, set a = (x² - n²)/y, b = 2nx/y, and c = (x² + n²)/y. A calculation shows these three numbers satisfy the two equations for a, b, and c above.

These two correspondences between (a,b,c) and (x,y) are inverses of each other, so we have a one-to-one correspondence between any solution of the two equations in a, b, and c and any solution of the equation in x and y with y nonzero. In particular, from the formulas in the two correspondences, for rational n we see that a, b, and c are rational if and only if the corresponding x and y are rational, and vice versa. (We also have that a, b, and c are all positive if and only if x and y are all positive; from the equation y² = x³ - xn² = x(x² - n²) we see that if x and y are positive then x² - n² must be positive, so the formula for a above is positive.)

Thus a positive rational number n is congruent if and only if the equation y² = x³ - n²x has a rational point with 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 y equal to 0, hence the existence of a rational point with 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

${\displaystyle N^{2}=ed^{2}+e^{2}}$

where

${\displaystyle {\begin{matrix}c&=&n^{2}/e+e\\a&=&2n\\b&=&n^{2}/e-e\end{matrix}}}$

Smallest solutions

The following is a list of the rational solution to ${\displaystyle a^{2}+b^{2}=c^{2}}$ and ${\displaystyle {\frac {ab}{2}}=n}$ with congruent number n and the smallest numerator for c. (we let a < b, note that a cannot be = b, because if so, then ${\displaystyle c={\sqrt {2}}a}$, but ${\displaystyle {\sqrt {2}}}$ is not rational number, thus c and a cannot be both rational numbers).

n	a	b	c
5	${\displaystyle {\frac {3}{2}}}$	${\displaystyle {\frac {20}{3}}}$	${\displaystyle {\frac {41}{6}}}$
6	3	4	5
7	${\displaystyle {\frac {35}{12}}}$	${\displaystyle {\frac {24}{5}}}$	${\displaystyle {\frac {337}{60}}}$
13	${\displaystyle {\frac {780}{323}}}$	${\displaystyle {\frac {323}{30}}}$	${\displaystyle {\frac {106921}{9690}}}$
14	${\displaystyle {\frac {8}{3}}}$	${\displaystyle {\frac {21}{2}}}$	${\displaystyle {\frac {65}{6}}}$
15	4	${\displaystyle {\frac {15}{2}}}$	${\displaystyle {\frac {17}{2}}}$
20	3	${\displaystyle {\frac {40}{3}}}$	${\displaystyle {\frac {41}{3}}}$
21	${\displaystyle {\frac {7}{2}}}$	12	${\displaystyle {\frac {25}{2}}}$
22	${\displaystyle {\frac {33}{35}}}$	${\displaystyle {\frac {140}{3}}}$	${\displaystyle {\frac {4901}{105}}}$
23	${\displaystyle {\frac {80155}{20748}}}$	${\displaystyle {\frac {41496}{3485}}}$	${\displaystyle {\frac {905141617}{72306780}}}$
24	6	8	10
28	${\displaystyle {\frac {35}{6}}}$	${\displaystyle {\frac {48}{5}}}$	${\displaystyle {\frac {337}{30}}}$
29	${\displaystyle {\frac {99}{910}}}$	${\displaystyle {\frac {52780}{99}}}$	${\displaystyle {\frac {48029801}{90090}}}$
30	5	12	13
31	${\displaystyle {\frac {720}{287}}}$	${\displaystyle {\frac {8897}{360}}}$	${\displaystyle {\frac {2566561}{103320}}}$
34	${\displaystyle {\frac {17}{6}}}$	24	${\displaystyle {\frac {145}{6}}}$
37	${\displaystyle {\frac {450660}{777923}}}$	${\displaystyle {\frac {777923}{6090}}}$	${\displaystyle {\frac {605170417321}{4737551070}}}$
38	${\displaystyle {\frac {1700}{279}}}$	${\displaystyle {\frac {5301}{425}}}$	${\displaystyle {\frac {1646021}{118575}}}$
39	${\displaystyle {\frac {5}{2}}}$	${\displaystyle {\frac {156}{5}}}$	${\displaystyle {\frac {313}{10}}}$
41	${\displaystyle {\frac {3280}{1023}}}$	${\displaystyle {\frac {1023}{40}}}$	${\displaystyle {\frac {1054721}{40920}}}$
45	${\displaystyle {\frac {9}{2}}}$	20	${\displaystyle {\frac {41}{2}}}$
46	${\displaystyle {\frac {253}{42}}}$	${\displaystyle {\frac {168}{11}}}$	${\displaystyle {\frac {7585}{462}}}$
47	${\displaystyle {\frac {11547216}{2097655}}}$	${\displaystyle {\frac {98589785}{5773608}}}$	${\displaystyle {\frac {217287944875297}{12111037689240}}}$
52	${\displaystyle {\frac {1560}{323}}}$	${\displaystyle {\frac {323}{15}}}$	${\displaystyle {\frac {106921}{4845}}}$
53	${\displaystyle {\frac {1472112483}{202332130}}}$	${\displaystyle {\frac {21447205780}{1472112483}}}$	${\displaystyle {\frac {4850493897329785961}{297855654284978790}}}$
54	9	12	15
55	${\displaystyle {\frac {1100}{117}}}$	${\displaystyle {\frac {117}{10}}}$	${\displaystyle {\frac {17561}{1170}}}$
56	${\displaystyle {\frac {16}{3}}}$	21	${\displaystyle {\frac {65}{3}}}$
60	8	15	17
61	${\displaystyle {\frac {6428003}{1423110}}}$	${\displaystyle {\frac {173619420}{6428003}}}$	${\displaystyle {\frac {250510625883241}{9147755349330}}}$
...	...	...	...
101	${\displaystyle {\frac {267980280100}{44538033219}}}$	${\displaystyle {\frac {44538033219}{1326635050}}}$	${\displaystyle {\frac {2015242462949760001961}{59085715926389725950}}}$
...	...	...	...
157	${\displaystyle {\frac {411340519227716149383203}{21666555693714761309610}}}$	${\displaystyle {\frac {6803298487826435051217540}{411340519227716149383203}}}$	${\displaystyle {\frac {224403517704336969924557513090674863160948472041}{8912332268928859588025535178967163570016480830}}}$

Current progress

Much work has been done classifying congruent numbers.

For example, it is known[6] that for a prime number p, the following holds:

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

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

Notes

1. Weisstein, Eric W. "Congruent Number". MathWorld.
2. Koblitz, Neal (1993), Introduction to Elliptic Curves and Modular Forms, New York: Springer-Verlag, p. 3, ISBN 0-387-97966-2
3. Ore, Øystein (2012), Number Theory and Its History, Courier Dover Corporation, pp. 202–203, ISBN 978-0-486-13643-1.
4. Conrad, Keith (Fall 2008), "The congruent number problem" (PDF), Harvard College Mathematical Review, 2 (2): 58–73, archived from the original (PDF) on 2013-01-20.
5. Darling, David (2004), The Universal Book of Mathematics: From Abracadabra to Zeno's Paradoxes, John Wiley & Sons, p. 77, ISBN 978-0-471-66700-1.
6. Paul Monsky (1990), "Mock Heegner Points and Congruent Numbers", Mathematische Zeitschrift, 204 (1): 45–67, doi:10.1007/BF02570859
7. Tian, Ye (2014), "Congruent numbers and Heegner points", Cambridge Journal of Mathematics, 2 (1): 117–161, arXiv:1210.8231, doi:10.4310/CJM.2014.v2.n1.a4, MR 3272014.

References

• Alter, Ronald (1980), "The Congruent Number Problem", American Mathematical Monthly, Mathematical Association of America, 87 (1): 43–45, doi:10.2307/2320381, JSTOR 2320381
• Chandrasekar, V. (1998), "The Congruent Number Problem" (PDF), Resonance, 3 (8): 33–45, doi:10.1007/BF02837344
• Dickson, Leonard Eugene (2005), "Chapter XVI", History of the Theory of Numbers, Dover Books on Mathematics, Volume II: Diophantine Analysis, Dover Publications, ISBN 978-0-486-44233-4 - see, for a history of the problem.
• Guy, Richard (2004), Unsolved Problems in Number Theory, Problem Books in Mathematics (Book 1) (3rd ed.), Springer, ISBN 978-0-387-20860-2, Zbl 1058.11001 - Many references are given it in.
• Tunnell, Jerrold B. (1983), "A classical Diophantine problem and modular forms of weight 3/2", Inventiones Mathematicae, 72 (2): 323–334, Bibcode:1983InMat..72..323T, doi:10.1007/BF01389327, hdl:10338.dmlcz/137483

External links

• Weisstein, Eric W. "Congruent Number". MathWorld.
• A short discussion of the current state of the problem with many references can be found in Alice Silverberg's Open Questions in Arithmetic Algebraic Geometry (Postscript).
• A Trillion Triangles - mathematicians have resolved the first one trillion cases (conditional on the Birch and Swinnerton-Dyer conjecture).