Robert Krzyzanowski Euler's Convenient Numbers Euler's C - PDF document

Robert Krzyzanowski Euler's Convenient Numbers Euler's Convenient Numbers§0. IntroductionIn the classical consideration of numbers of the form , we find a set of
convenient formulas for certain values of . Recall in Cox (see [1, 2.28]) a list is given ofcongruences for primes :Proposition 0.1. If is a prime, then
 \n
mod \r
\n \n\n\n\n
mod \r
\n \n\n\r
mod 
\n \n\n\n\r
mod 
\n \n
mod \r
 \n\n\n\r\r\n\n
mod 
 \n\n\r
mod \nCertain numbers , as we shall see, yield a nice set of congruences. Specifically,numbers such that if is odd, , and has only one solution\n

for non-negative , then is prime. It turns out is not such a number. For example,
\n\nusing the previous condition, is the only representation of as ,\n\n\n\n
\n\nyet is not prime. If we wish to find primes such that , then we would\n
\n\nhave to use the methods of class field theory (as developed in Cox, [1, §5-9]).The numbers that do provide convenient congruences are aptly named convenientnumbers. Although class field theory makes the existence of these numbers less crucialfor computing primes of the form , they are still interesting for historical and
computational reasons. For example, Euler was able to find the prime\n\n\n\n\r\n,,by noticing is convenient. In this paper, we will see that there are only finitely many\n\rof these numbers. In fact, the highest such number is -- the Riemann hypothesis\n\rifholds! If it does not, there is at most one more (pretty large) such number. This last fact isa deep result due to Weinberger that we will sketch at the end. First, we recall some basictheorems and facts from genus theory.§1. Genus TheoryProposition 1.1. Let be an integer and be an odd integer relatively\n
mod 4prime to . Then is properly represented by a primitive form of discriminant if andonly if is a quadratic residue modulo . As a corollary, if is an integer and is an odd prime not dividing , then if and only if is represented by a primitive\nform of discriminant .\r Robert Krzyzanowski Euler's Convenient NumbersProof. See [1, Lemma 2.5] and [1, Corollary 2.6] in Cox. Theorem 1.2. (Classification of Primitive Positive Definite Forms) Let \n
mod with , and let be the set of primitive positive definite forms of
discriminant . Then Dirichlet composition induces a well-defined binary operation on

, which makes into a finite Abelian group whose order is the class number

. Furthermore, the identity element of is the class containing the principalform
 if ,
mod

\n\n if
mod and the inverse of the class containing the form is the class containing



Proof. See [1, Theorem 3.9] in Cox. Definition 1.3. The group in the previous theorem is called the . The
class groupprincipal form of discriminant is called the The form 

principal class. is called the of so that the opposite form gives the inverseopposite 

under Dirichlet composition.Proposition 1.4. A reduced form of discriminant has



order in the class group if and only if or .!
Proof. See [1, Lemma 3.10] in Cox. Proposition 1.5. Let with . Take to be the number of odd\n"
mod primes dividing . Define the number as follows: if , and if\n"


mod , then \r# , then , where , and is given by:
mod ""\n\n"\n\r"if if if if



mod mod mod mod .Then the class group has exactly elements of order .!
\nProof. See [1, Proposition 3.11] in Cox. Theorem 1.6. (Main Theorem of Genus Theory) mod Let with .\n
Then(i) All genera of forms of discriminant consist of the same number of classes.(ii) There are genera of forms of discriminant , where is given in the\nprevious proposition. Robert Krzyzanowski Euler's Convenient Numbers(iii) The principal genus consists of the classes in , the subgroup of squares in
the class group 
Proof. See [1, Theorem 3.15] in Cox. Proposition 1.7. Let and be primitive forms of discriminant ,
$
%

positive definite if . Then the following statements are equivalent:(i)
$


 and are in the same genus, i.e., they represent the same values in
&'(ii)
$
&


 and represent the same values in for all non-zerointegers .(iii)

 and are equivalent modulo for all non-zero integers .$

Proof. See [1, Theorem 3.21] in Cox. §2. Convenient numbersIn §0, we introduced convenient numbers: those for which genus theory gives acongruence condition for . In this section, we will pass to the language of classes
using our knowledge from §. Notice that investigating convenient numbers is the sameas considering each genus of discriminant that consists of a single class. The\rfollowing theorems make this precise.Theorem 2.1. Let be a positive integer. Then the following statements are equivalent:(i) Every genus of forms of discriminant consists of a single class.\r(ii) If is a reduced form of discriminant , then either ,

\r, or .(iii) Two forms of discriminant are equivalent if and only if they are properly\requivalent.(iv) The class group is isomorphic to for some integer .\r&

(v) The class number equals , where is as in Proposition 1.5.\r
\nProof. Let be the class group for discriminant . Assume each genus\r\r
of forms of discriminant consists of a single class. We will show that if\r

\r is a reduced form of discriminant , then either , , or. From Proposition 1.4, it suffices to show that a reduced form of discriminant has order or in . From Theorem 1.6(iii) (the main theorem of genus theory), we\nknow that the principal genus is . Since we assumed each genus of forms ofdiscriminant consists of a single class, by definition this gives . Hence,\r\neach element in has order . This shows (i)(ii).!Assume that if is a reduced form of discriminant , then either

\r\r, , or . We will show two forms of discriminant are equivalent ifand only if they are properly equivalent. The right-to-left implication is obvious. Withoutloss of generality, consider two equivalent reduced forms of discriminant , say\r


$
'
)
 

 and . Robert Krzyzanowski Euler's Convenient NumbersThen by definition there are so that*"+,
$
*"
+

with . First, notice we can use our assumption to give , or+*"-\n $ . We claim that is properly equivalent to or its opposite. If is not properlyequivalent to , it must be improperly equivalent to , that is, . But thenif$$+*"\nwe let denote the opposite of ,$$$
*"
+'
*)
*"
+ "
+'
*)
*"
+ "
+$
*"
+











with . Hence, is properly equivalent to the



+*"+*"\n\n opposite of . This proves the claim. We continue by showing that reduced forms areproperly equivalent to their opposite (so in particular, and are properly equivalent to $their opposite). Let be a reduced form and its opposite.If



., then this is trivial. If , then













.




so that since , by definition is properly equivalent to . Similarly,\n\n\n\n
if , then








.


so that again is properly equivalent to . Hence, we have shown that the original and are properly equivalent to each other or their opposite, but since they are reduced, thatmeans their opposites are properly equivalent to each other, and hence and are $properly equivalent. This shows (ii)(iii).Assume two forms of discriminant are equivalent if and only if they are\rproperly equivalent. We want to show the class group for some .0&,
It is easy to see any form is equivalent to its opposite through . Hence,


1
by our assumption, for any form in its opposite must lie in as well since they areproperly equivalent. Then the last statement from Theorem 1.2 tells us the opposite of aform and its inverse are in the same class in , so that each class is its own inverse. Inother words, is a finite abelian group with all elements of order . Then by theclassification of finite abelian groups, for some (since a0&,
component of any other for would give an element of order ). This shows&''#'that (iii)(iv).Furthermore, assume for some . By Theorem 1.6(ii), the0&,
number of genera is the index of in , that is, . Hence, the class number\n\r2
 \n \nSince we assumed , obviously 1. Hence, . This0&0\r

\nprove (iv)(v). Robert Krzyzanowski Euler's Convenient NumbersFinally, assume . Then as above, means that\r\r


\n\n0\n, so that by Theorem 1.6(iii), the principal genus consists of a single class.However, the genera of forms all consist of the same number of classes, so each genusconsists of a single class. This means (v)(i), which concludes the proof. We will now make the connection between convenient numbers and forms ofdiscriminant whose genus consists of a single class. For sake of preciness, we will\ruse Euler's traditional definition of a convenient number.Definition 2.2 Let be an odd number relatively prime to , which is properlyrepresented by . If the equation has only one solution with


3, then is a prime number, and is called a .convenient numberProposition 2.3. A positive integer is a convenient number if and only if for forms ofdiscriminant , every genus consists of a single class.\rThe above proposition asserts the in Theorem 2.1 can be given by all five equivalentdefinitions (i)-(v).Lemma 2.4. Let be a positive odd number relatively prime to . Then the#\nnumber of ways that is properly represented by a reduced form of discriminant is\r \n .\r\r4Proof. Let and be odd with . Let be a prime dividing . We#\n#\n
will show that mod has

 \r\r4\nsolutions. Recall that a polynomial
"

mod has solutions, where is the number of solutions of mod with

55"" the prime decomposition of . (This is a direct application of the ChineseRemainder Theorem; for a detailed proof, see [12, Theorem 5.25] in Apostol). In otherwords, to find the number of solutions for

mod ,it is sufficient to know the number of solutions of
mod ,where is a component of the prime decomposition of . However, notice5'5 that if , then this last equation has no solutions, so indeed it has \n\n \n solutions. Now consider . This means there is at least one solution.Assume there were multiple solutions, that is, there is a such that
4&mod (with ). Robert Krzyzanowski Euler's Convenient NumbersThen divides . However, it can only divide one of the two




factors since and (i.e., it doesn't divide their sum). But




4
then
-mod so indeed there are only two solutions. Again, there are solutions.\n\rHence, in total, there are \r\r4\nsolutions to the congruence mod .

Now, consider forms of discriminant of the form$
\r
$


!(


 .Then the discriminant of is . Hence, we need .$
\r\r

However, since and are fixed, notice determines uniquely. Hence, this isprecisely equivalent to solutions for
mod , that is,
mod .In other words, there is a bijection between and solutions of mod .$



Take to be a form of discriminant and let


\r
 67"+6+7"\n
 be a proper representation. Let be such that , and let""68++788,6+7"\n and . Then as varies, we get all solutions of .Finally, let$
 6
"7
+

Now$
6
"6
"7
+7
+




6
6"
"67
6+
"7
"+7
7+
+6677
6"6+"77+
""++



6"7+6+7"



Now, pick the unique so that . Then and so8,6+7"!6"7+$
($



 is uniquely represented as . Call this . Notice that the map sending a67proper representation to is onto, since each of the above steps is 67$


67reversible (so that we can start with and show there is an such that$

67

$
 6
"7
+67

).Assume with$
$
6767..

 Robert Krzyzanowski Euler's Convenient Numbers6767"+"+....\nUsing the fact (by determinants) and expanding6+"7&6+"7

....







, we see that . Then by the sameargument as in [1, Theorem 2.8] in Cox (assuming is reduced), .
-
\n\nFinally, this implies if and only if (just$
$
67-676767....



multiply by on the left in the above matrix equation), so that the map sending a67"+....proper representation to the form is two-to-one in addition to being 67$


67onto. Hence, by the earlier correspondence between forms as in and solutions of


mod , there are twice as many ways to properly represent by a reducedform of discriminant as there are solutions to mod . We have already\r

computed the latter, so indeed there are 4\r\r\nrepresentations in total. Corollary 2.5. Let be properly represented by a primitive positive definite form
\r#\n
 of discriminant 768 and assume that is odd and relatively primeto . If denotes the number of prime divisors of , then is properly represented in"exactly ways by a reduced form in the genus of .
"\n
Proof. Since two forms representing do not have disjoint values in , they&\r
 must lie in the same genus. Then by Proposition 1.1, for each prime , 4\n(since implies ), so that by the previous lemma is properly represented
\n4in ways. ""\nProof. (of Proposition 2.3) We will first show the right-to-left implication. Assume thatfor forms of discriminant , every genus consists of a single class. By Definition 2.2,\rwe need to show that if is properly represented by and 

(), then is prime. Since is the only reduced form in its genus, by the
3
previous Corollary (2.5), is properly represented by in ways. Now, since
"\n
\n&\r can be positive or negative and can be positive or negative, at least of these arethe ones with and both positive. That is, there are ways of writing

"\nwith . We assumed that has a unique such solution, so that , that is,
3\n"\n"\n8,. In other words, has a single prime divisor, so that for some . If88\n83, then is already prime. If (i.e., is not a prime), then by Lemma 2.4, 8has at least representations. But then has at least representations so isproperly represented in at least ways with which contradicts our assumption
3that it is represented in such a way uniquely. Hence, gives a contradiction, so that83indeed is prime. By Definition 2.2., is hence a convenient number.Conversely, assume is convenient. Take to be a form of discriminant ,
\r
and let be the Dirichlet composition of with itself. Without loss of$



generality, assume that is reduced. Then by Theorem 2.1, if $
$



then since each element in the class group has order , each genus consists of a single! Robert Krzyzanowski Euler's Convenient Numbersclass. Hence, it suffices to show . By way of contradiction, assume$


that . Assume are odd primes not dividing which are$
%
%*
represented by . Then since is the Dirichlet composition of with
$




itself, is represented by . Then by Corollary 2.5, since , has *
"*
\nproper representations by a reduced forms of discriminant . Since of these is due to\r\n$
*

3
, there are at most ways to write . If satisfy this equation,then so do , and . Hence, if there were two unique solutions for





*

3 with , then there would be in total, a contradiction. Hence,*

3*\n* has a unique solution for . Then since is properly
represented by , and has a unique solution for , by
*

3Definition 2.2, should be a prime since is convenient. Of course, it isn't, which gives*the desired contradiction. Therefore, . $


In this section, we looked at convenient numbers more carefully, and showed theircharacterization by the statements in Theorem 2.1. We will now look concretely at whichnumbers are convenient.§3. Existence of only finitely many convenient numbersIn the previous section, we used genus theory to talk about the connection betweenconvenient numbers and forms of discriminant whose genus consists of a single\rclass. Convenient numbers allow us to provide an elementary condition (congruence) forwhen a number is prime. Hence, we conclude by investigating which numbers
are convenient. It can be checked manually that the following integers are!\n\rconvenient numbers.Table 3.1. List of convenient numbers .!\n\r \r\n\r\n\n\n\n\n\n\r\n\r\r\r\r\r
all such with one class per genus\n\n\n\n\n\n\n\n\n\n\n\n\r\n\r\r\r\n\r\n\n\n\rEuler and Gauss noticed early on that there do not seem to be any more convenientnumbers immediately after 1848. In 1914, D. N. Lehmer verified this up to ,,\nusing sieve methods and an electro-mechanical computer (for more on this, read about theLehmer sieve). Euler was very troubled by the sudden disappearance of convenientnumbers past 1848 (the existence of infinitely many such numbers would make classicalmethods for finding large primes much more effective). In order to convince himself thatthere were indeed no such relatively small numbers past 1848, Euler proved the followingproposition. This helped him narrow down which numbers could be convenient.Proposition 3.2. Let be a convenient number. Then Robert Krzyzanowski Euler's Convenient Numbers(i) If is a perfect square, then or .\n\r(ii) mod If , then is convenient.\r
(iii) mod If is convenient and , then is convenient.\r\r
(iv) If so that is convenient, then is convenient.8,8(v) mod If , then is convenient.
(vi) mod If with , then is not convenient.#\n\n\r
(vii) mod If , then is convenient.\r
(viii) mod If , then is not convenient.\r
\n(ix) mod If , then is not convenient.\n\r
(x) If there is a prime with such that , then is not,\r\rconvenient.Proof. Euler attempted to prove all of these in [3], but (iv), (vi), (viii), and (ix) haderrors. Grube corrected these in [2], completing the proof of the theorem. If one spendstime looking at these conditions, then it becomes increasingly clear that the convenientnumbers significantly thin out at some point (e.g., 1848). This proposition certainly provides some insight into the nature of convenientnumbers, but it does not give any information about how many there are. This did notbecome more apparent until 1934, when S. Chowla proved ([4]) that there are finitelymany convenient numbers. This paper used the fact that lim'9:'$'

:where denotes the number of genera of binary quadratic forms with discriminant .$''
In 1954, Briggs and Chowla used Siegel's asymptotic formula to give some concretebounds (see [5]). Later, in 1963, E. Grosswald improved on this further by using theanalytic class number formula (see [6]), and all this culminated in the most current resultby Weinberger (see [7]), which states there are no convenient numbers past unless a\n\rweaker assumption than the Riemann hypothesis is false, in which case there is at mostone.Theorem 3.3. There is at most one convenient number larger than .\n\rThe proof of this theorem uses some deep results discovered by Tatuzawa, and theproof of these is beyond the scope of this paper. We begin with a reminder of the notationand nature of L-series. First, the special case
+\nis the Riemann zeta function. This function has no zeros when Re, and it has
+#\ntrivial zeros for even negative numbers. Finally, any non-trivial zero lies in the strip
+,4+\nRe, the "critical strip." The Riemann hypothesis states that any non-trivial zero has Re; in other words, the only zeros of are the negative

++ Robert Krzyzanowski Euler's Convenient Numbers10 even numbers and those with real part . It is obvious that lemmas 3.4 and 3.5 give aweaker criterion, and hence are not as strong as the Riemann hypothesis.The generalization of is the Dirichlet L-series. If is a Dirichlet character, then\n
 ;+
"\n
for each complex number with Re. The zeros of the L-series shares the+,+#\n
same characteristics as the Riemann zeta function. Indeed, the generalized Riemann hypothesis states that in the critical strip, any satisfies Re. In our;++

case, we will use \n\n

""''the Kronecker symbol (see [1, pg. 104] and [1, §7.D] in Cox). In this case, ;

"\n'Furthermore, assume (so that we are excluding ). For a quadratic field'\r#\r\n 8'')'

, we denote the class number by . Furthermore, define be theleast positive integer such that for each in the ideal class group\n)'
\n
(where is understood as the ideal class of the principal ideals, which formsan identity on ). This number is called the exponent of the ideal class group.)'

Then if , by definition each element has order , so by Theorem 2.1)',!
these () are exactly the convenient numbers (where we have preserved some'\rresults using ideals instead of forms--see the last paragraph on [1, pg. 112 and Theorem5.30 and Theorem 7.22], which states for , ). Finally, we assume''0

88the aforementioned lemmas. Lemma 3.4. For fixed , if and , then has a'3);\n!;+\r\n\n='\n&

real root , with .+\n&\r!+\nLemma 3.5. maxFor fixed , there is at most one with and#''3))\n&\n\n= ;\n!
=Proof. These lemmas are proved in Tatuzawa [17]. We can now start proving the theorem.Notation. Let be the product of the first primes (starting at ).'" Lemma 3.6. If in the interval with , then;+%\n!+\n'3'
\n\nlog )'#'3''3'
 for all . Without this hypothesis, there is at most one such\n\n\n\nthat .)'
Proof. As noted earlier, if , then is not going to be a convenient number. In)'#
other words, this is saying there are no convenient numbers larger than . It has already\n\nbeen checked computationally by D. H. Lehmer and J. D. Swift that there are noconvenient numbers up to Robert Krzyzanowski Euler's Convenient Numbers11'\r\n�\n\n\n\n\n,,,.First, from Theorem 2.1, when . If , then since is')'3\n\n

\nthe th prime, is the product of (the first 11 primes) and subsequent primes larger\n\n''\n\nthan (and the product of these latter primes is obviously ). Then\n\n3\n\ncombined with the above approximation of , we have\n\n'3'3\n==

\n\n\n\n\n\n\n\n
Furthermore, when (see [1, §7.D] in Cox). It is a result from algebraic'3')'
number theory (see [14, Cohen], [15, Cohen], [16, Cohn], and a few hints in [1, §7.D] inCox) that '
;\n8'

where 8''#'
'?''ln

for ,for is the Dirichlet structure constantwith the fundamental unit and is the number

'?'of substitutions that leave the binary quadratic form unchanged, given by?''\r'\r
for for otherwise. Since we assumed and (so that and ),''#\r?'8'&'&'

 '=';\n


This is important as it gives a relationship between the class number and .';\n

As we will see, it is precisely this estimation of that will yield a contradiction for'
)'
 We will use the contrapositive of Lemma 3.4 with . We log assumed;+%
 in , so if that would give a contradiction. Hence,\n&\r!+!\n;\n!\r\n
= ;\n#
=Then from and this fact,
= ';\n#\n

'=='\n''='=)')'3�\n\n=#\n\nln ln ln \n&'\n\n\n\nln Hence, we need . Now, we can keep going using (3.),#\n\n Robert Krzyzanowski Euler's Convenient Numbers12 #\n='='=)'))'\n\n#''=)'\n\n'




ln ln ln lnln \n\n\n\n\n\n\n\n\n\n\n\n&\n\n\n\n\n\n\n\n


 ln ln ln )'\n\n='

\n\n\n\n\n\n\n\nDividing both sides by , we get\n\n \n\n\n\n#=)'\n\n'
#\n\n
\n\n\n\n\n\nln ln since obviously ==)'\n\n)'''
#
\n\n\n\n\n\n\n\n\n\nln ln ln so that we can use the lower bound found earlier. However, , so\n\n\n\r#&\n\n\nthat our assumption cannot be true. Hence, for .)')'#'3'

\n\n Now assume we do not have the Riemann hypothesis. This time, we let butln \n\n for Lemma 3.5. If ;!=, we don't have a contradiction. Otherwise, exclude the'3))'max\n&\n\n=\n\n in the lemma from consideration. Then, again assume so we can say)'
 ';\n#\n

==='')'''\r\r3#\n\n\n\n\n\n\n\n\n\nln so that . Now, using (3.7),#\n\n '3'='=''#==)')'''=\r

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n&\n\n

\r\rln ln \n\n'\n\nln
 n ln ')'##\n\n=\n\n\n\n\n\n\n\n\n\n\n\n


Then, again, \n\n\n\n\n\r#\n\nwhich is a contradiction, so that for all except at one most one. In)'#'3'
\n\nother words, there are no convenient numbers beyond when the Riemann hypothesis\n\nholds, and if it does not then there is at most one. Once again, it has been verifiedcomputationally that is the biggest convenient number . Therefore, this\n\r!'\n\nconcludes the proof of Theorem 3.3. References[1] Cox, D. Wiley, 1989."Primes of the form ".
 Robert Krzyzanowski Euler's Convenient Numbers13[2] Grube, F. Ueber einige Eulersche Sätze aus der Theorie der quadratischen Formen.Zeitschrift für Mathematik und Physik 19 (1874), pp. 492-519.[3] Euleri, L. De formulis specei ad numeros primos explorandos idoneis

earumque mirabilibus proprietatibus. Opera Omnia, Series I, Vol. 4 (CommentationesArithmeticae III), pp. 269-289.[4] Chowla, S. Quarterly Journal ofAn extension of Heilbronn's Class Number Theorem. Mathematics (Oxford) 5 (1934), pp 304-307.[5] Chowla, S. and Briggs, W. E. On discriminants of binary quadratic forms with asingle class in each genus. Canadian Journal of Mathematics 6 (1954), pp 463-470.[6] Grosswald, E. Negative discriminants of binary quadratic forms with one class ineach genus. Acta Arithmetica 8 (1963), pp. 295-306.[7] Weinberger, P. J. . ActaExponents of the class groups of complex quadratic fieldsArithmetica 22 (1973), pp. 117-124.[8] Cohen, H. . New York:A Course in Computational Algebraic Number TheorySpringer-Verlag, 1993.[9] Cohen, H. . New York: Springer-Advanced Topics in Computational Number TheoryVerlag, 2000.[10] Cohn, H. Advanced Number Theory. New York: Dover, 1980.[11] Tatuzawa, T. . Japanese Journal of Mathematics. 21 (1951),On a theorem of Siegelpp. 163-178.[12] Apostol, T. New York: Springer-Verlag,Introduction to Analytic Number Theory.1976.