Heegner number

In number theory, a Heegner number is a (square-free) positive integer "d" such that the imaginary quadratic field$mathbf\{Q\}(sqrt\{-d\})$has class number 1. Equivalently, its ring of integers has a unique factorization.

The determination of such numbers is a special case of the class number problem, and they underlie several striking results in number theory.

According to the Stark–Heegner theorem there are precisely nine Heegner numbers::num|1, num|2, num|3, num|7, num|11, num|19, num|43, num|67, num|163.This result was conjectured by Gauss and proven by Kurt Heegner in 1952.

Euler's prime-generating polynomial

Euler's prime-generating polynomial:$n^2\; +\; n\; +\; 41,$which gives (distinct) primes for $n=0,dots,39$, is related to the Heegner number $163\; =\; 4\; cdot\; 41\; -\; 1$.

Rabinowitz [Rabinowitz, G. "Eindeutigkeit der Zerlegung in Primzahlfaktoren in quadratischen Zahlkörpern." Proc. Fifth Internat. Congress Math. (Cambridge) 1, 418-421, 1913.] proved that:$n^2\; +\; n\; +\; p$gives primes for $n=0,dots,p-2$ if and only if its discriminant $1-4p$ is a Heegner number.

(Note that $p-1$ yields $p^2$, so $p-2$ is maximal.)1, 2, and 3 are not of the required form, so the Heegner numbers that work are $7,\; 11,\; 19,\; 43,\; 67,\; 163$, yielding prime generating functions of Euler's form for $2,3,5,11,17,41$; these latter numbers are called "lucky numbers of Euler" by F. Le Lionnais. [Le Lionnais, F. Les nombres remarquables. Paris: Hermann, pp. 88 and 144, 1983.]

Almost integers and Ramanujan's constant

Ramanujan's constant is the transcendental number Fact|date=June 2008 $e^\{pi\; sqrt\{163$, which is an almost integer, in that it is very close to an integer:

:$e^\{pi\; sqrt\{163\; =\; 262,537,412,640,768,743.999\; 999\; 999\; 999\; 25ldots$ [ [http://mathworld.wolfram.com/RamanujanConstant.html Ramanujan Constant - from Wolfram MathWorld ] ] $approx\; 640,320^3+744$

This coincidence is explained by complex multiplication and the "q"-expansion of the j-invariant.

Detail

Briefly, $j((1+sqrt\{-d\})/2)$ is an integer for "d" a Heegner number, and $e^\{pi\; sqrt\{d\; approx\; -j((1+sqrt\{-d\})/2)\; +\; 744$ via the "q"-expansion.

If $au$ is a quadratic irrational, then the "j"-invariant is an algebraic integer of degree $|mbox\{Cl\}(mathbf\{Q\}(\; au))|$, the class number of $mathbf\{Q\}(\; au)$ and the minimal (monic integral) polynomial it satisfies is called the Hilbert class polynomial.

Thus if the imaginary quadratic extension $mathbf\{Q\}(\; au)$ has class number 1 (so $d$ is a Heegner number), the "j"-invariant is an integer.

The "q"-expansion of "j", with its Fourier series expansion written as a Laurent series in terms of $q=exp(2\; pi\; i\; au)$, begins as::$j(q)\; =\; frac\{1\}\{q\}\; +\; 744\; +\; 196,884\; q\; +\; cdots$The coefficients $c\_n$ asymptotically grow as $ln(c\_n)\; sim\; 4pi\; sqrt\{n\}\; +\; O(ln(n))$, and the low order coefficients grow more slowly than $200,000^n$, so for $q\; gg\; 200,000$, "j" is very well approximated by its first two terms.

Setting $au\; =\; (1+sqrt\{-163\})/2$ yields $q=-exp(-pi\; sqrt\{163\})$ or equivalently, $frac\{1\}\{q\}=-exp(pi\; sqrt\{163\})$.Now $j((1+sqrt\{-163\})/2)=(-640,320)^3$, so:$(-640,320)^3=e^\{pi\; sqrt\{163+744+O(e^\{-pi\; sqrt\{163)$Or,:$e^\{pi\; sqrt\{163=640,320^3+744+O(e^\{-pi\; sqrt\{163)$where the linear term of the error is:$-196,884/e^\{pi\; sqrt\{163\; approx\; 196,884/(640,320^3+744)approx\; -.00000000000075$explaining why $e^\{pi\; sqrt\{163$ is within approximately the above of being an integer.

Other Heegner numbers

For the four largest Heegner numbers, the approximations one obtains [These can be checked by computing $sqrt\; [3]\; \{e^\{pisqrt\{d-744\}$ on a calculator, and$196,884/e^\{pisqrt\{d$ for the linear term of the error.] are as follows.:Alternatively [http://groups.google.com.ph/group/sci.math.research/browse_thread/thread/3d24137c9a860893?hl=en#] ,:where the reason for the squares is due to certain Eisenstein series. For Heegner numbers , $196,884/e^\{pisqrt\{d\; >\; 1$, so one does not obtain an almost integer; even $d=19$ is not noteworthy.

The "j"-invariants associated to the cubes above are highly factorable and factor as:

For $d=3$, the fit is reasonably good, but does not follow from the above::$e^\{pi\; sqrt\{3\; approx\; (-2^3)^3+744-1.2$

Consecutive primes

Given an odd prime $p$, if one computes $k^2\; pmod\{p\}$ for $k=0,1,dots,(p-1)/2$ (this is sufficient because $(p-k)^2equiv\; k^2\; pmod\{p\}$), one gets consecutive composites, followed by consecutive primes, if and only if $p$ is a Heegner number. [http://www.mathpages.com/home/kmath263.htm]

For details, see "Quadratic Polynomials Producing Consecutive Distinct Primes and Class Groups of Complex Quadratic Fields" by Richard Mollin.

References

External links

*
* [http://www.research.att.com/~njas/sequences/?Anum=A003173 A003173] from the On-Line Encyclopedia of Integer Sequences
* [http://www.ams.org/bull/1985-13-01/S0273-0979-1985-15352-2/S0273-0979-1985-15352-2.pdf Gauss' Class Number Problem for Imaginary Quadratic Fields, by Dorian Goldfeld] : Detailed history of problem.