Euler's φFunction on Arithmetic ProgressionsAuthor(s): D. J. NewmanSource: The American Mathematical Monthly, Vol. 104, No. 3 (Mar., 1997), pp. 256-257Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2974791 .

Accessed: 10/12/2014 09:53

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org.

.

Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toThe American Mathematical Monthly.

http://www.jstor.org

This content downloaded from 128.235.251.160 on Wed, 10 Dec 2014 09:53:55 AMAll use subject to JSTOR Terms and Conditions

NOTES Edited by Jimmie D. Lawson

Euler's + Function on Arithmetic Progressions

D. J. Newman

We will study the + function's behavior on arithmetic progressions. The numerical evidence is extremely misleading in many cases. Thus it was pointed out by Dov Jarden in his book Recumng Sequences, Riveon Lematematika, Jerusalem, 1973 page 65, that +(30n + 1) > +(30n) for all n < 10000. (Indeed, further computa- tions have shown that this persists up to 20,000,000.) Another case that resists is that of 6n + 1 vs. 6n + 2, where again, the inequality +(6n + 1) > +(6n + 2) holds well into the millions.

At least, in this case, we were able to explicitly produce the smallest n predicted by our theorem: it is 6,197,024. In Jarden's case and in many others, the n is not explicitly available and it may be beyond the reach of any possible computers! At any rate we can prove the following.

Theorem. If a, b, c, d are nonnegative integers with a, c > 0 and ad - bc + 0 then there exists a positive integer n for which +(an + b) < +(cn + d).

Remarks: By replacing b by b + aN and d by d + cN, N large, we see in fact that our theorem gives infinitely many such n. The condition ad - bc + 0 is certainly a necessary one since otherwise we would have the case of a = c, b = d which gives equality of +(an + b) and +(cn + d), but even worse we would have the case of +(4n) which is always strictly bigger than +(n).

Proof of the theorem: Begin with the linear Diophantine equation ax + b = Py, where P is chosen prime to a and with (+(P)/P) < , e later to be specified. For example, P may be chosen as the product of many consecutive primes. Notice that then

+(P) (t(ax + b) = +(Py) < +(P)y = p (ax + b) < e(ax + b).

Next observe that a general solution to our equation is given by x = xO + kP, y = yO + ka, k an arbitrary integer. This gives cx + d = cxO + d + kcP, and if we denote 8 = gcd(cxo + d, cP) and note that it divides a(cxO + d) - yOcP = a(cxO +

d) - c(axO + b) = ad - bc, we obtain 8 < lad - bel. Factoring out 8 gives cx + d = b(A + kB) where A and B are relatively prime.

We now recall Dirichlet's great theorem on primes in arithmetic progressions. This states that if A and B are relatively prime integers with B > 0, then there are infinitely many integers, k, for which A + Bk is a prime. We may apply Dirichlet's theorem to our case and thereby choose k so as to make A + Bk a prime, which we shall call p.

[March 256 NOTES

This content downloaded from 128.235.251.160 on Wed, 10 Dec 2014 09:53:55 AMAll use subject to JSTOR Terms and Conditions

Now we have Now we have

cx + d +(cx + d ) = +( 8p) > 0( p) - p - 1 2 p/2 = 2^

Since (cx + d)/(ax + b) is monotonic and positive, it has a positive minimum, m, so

cx + d m(ax + b) m(ax + b)

2bi 26 2lad - bcl ( )'

where e is defined by this last equation.

1830 Rittenhouse Square, Philadelphia, PA 19103

| I _ . . .. I . | . | .1

A Note on Weyl's Inequality

Steve Fisk _ I .. . I

We present a simple inequality for the eigenvalues of Hermitian matrices that implies Weyl's inequality and the monotonicity theorem. The underlying idea, to intersect suitably chosen subspaces to obtain eigenvalue inequalities, is not new. See [1] and [2].

Lemma 1. If 51, . . . S Sk are subspaces of an n-dimensional vector space X, and if dim(S1) + *** +dim(Sk) > n(k - 1), then the intersection of all the Si's is non-zero.

Proof: Consider the map from the direct sum of the subspaces Si to the sum of k - 1 copies of V that sends (v1, . . ., vk) to (V1 - V2, V2 - V3, * * *, Vk - Vk-l )* The intersection of all the Si's is the kernel of this map, and has dimension at least dim(S1) + *** +dim(Sk) - (k- 1)n. -

If H is an n-by-n Hermitian matrix, we denote its ordered eigenvalues by A1(H) < * < An(H). An n-by-n matrix X is negative semidefinite if v*Xiv < O for every vector v. For instance, the zero matrix is negative semidefinite.

Theorem 1. Suppose H1,..., Hk are n-by-n Helimitian matrices such that Hl + H2 + *- * +Hk is negative semidefinite. Then

Ail( H1 ) + Ai2( H2 ) + * * * + Aik( Hk ) < °

foralli1,...,ik E {1,...,n}suchthati1 + *** +ik vn +k.

Proof: Let Sj be the subspace spanned by eigenvectors of Hj corresponding to the eigenvalues Aij(Hj), Aij+l(Hj), . . . S An(Hj). Since

k k

E dim(Sj) = , (n - ij + 1) = nk + k - (i1 + *** +ik) > n(k - 1), j-l j=l

Lemma 1 ensures that there is a unit vector x in the intersection of all the Sj's.

cx + d +(cx + d ) = +( 8p) > 0( p) - p - 1 2 p/2 = 2^

Since (cx + d)/(ax + b) is monotonic and positive, it has a positive minimum, m, so

cx + d m(ax + b) m(ax + b)

2bi 26 2lad - bcl ( )'

where e is defined by this last equation.

1830 Rittenhouse Square, Philadelphia, PA 19103

| I _ . . .. I . | . | .1

A Note on Weyl's Inequality

Steve Fisk _ I .. . I

We present a simple inequality for the eigenvalues of Hermitian matrices that implies Weyl's inequality and the monotonicity theorem. The underlying idea, to intersect suitably chosen subspaces to obtain eigenvalue inequalities, is not new. See [1] and [2].

Lemma 1. If 51, . . . S Sk are subspaces of an n-dimensional vector space X, and if dim(S1) + *** +dim(Sk) > n(k - 1), then the intersection of all the Si's is non-zero.

Proof: Consider the map from the direct sum of the subspaces Si to the sum of k - 1 copies of V that sends (v1, . . ., vk) to (V1 - V2, V2 - V3, * * *, Vk - Vk-l )* The intersection of all the Si's is the kernel of this map, and has dimension at least dim(S1) + *** +dim(Sk) - (k- 1)n. -

If H is an n-by-n Hermitian matrix, we denote its ordered eigenvalues by A1(H) < * < An(H). An n-by-n matrix X is negative semidefinite if v*Xiv < O for every vector v. For instance, the zero matrix is negative semidefinite.

Theorem 1. Suppose H1,..., Hk are n-by-n Helimitian matrices such that Hl + H2 + *- * +Hk is negative semidefinite. Then

Ail( H1 ) + Ai2( H2 ) + * * * + Aik( Hk ) < °

foralli1,...,ik E {1,...,n}suchthati1 + *** +ik vn +k.

Proof: Let Sj be the subspace spanned by eigenvectors of Hj corresponding to the eigenvalues Aij(Hj), Aij+l(Hj), . . . S An(Hj). Since

k k

E dim(Sj) = , (n - ij + 1) = nk + k - (i1 + *** +ik) > n(k - 1), j-l j=l

Lemma 1 ensures that there is a unit vector x in the intersection of all the Sj's.

1997] 1997] 257 257 NOTES NOTES

This content downloaded from 128.235.251.160 on Wed, 10 Dec 2014 09:53:55 AMAll use subject to JSTOR Terms and Conditions