Any n Arithmetic Progressions Covering the First 2nIntegers Cover All Integers

Any n Arithmetic Progressions Covering the First 2n Integers Cover All IntegersAuthor(s): R. B. Crittenden and C. L. Vanden EyndenSource: Proceedings of the American Mathematical Society, Vol. 24, No. 3 (Mar., 1970), pp.475-481Published by: American Mathematical SocietyStable URL: http://www.jstor.org/stable/2037391 .

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ANY n ARITHMETIC PROGRESSIONS COVERING THE FIRST 2n INTEGERS COVER ALL INTEGERS

R. B. CRITTENDEN AND C. L. VANDEN EYNDEN1

In 1958 S. Stein [7] defined a system of n congruences x-ai (mod b,), in, to be disjoint if no x satisfies more than one of them. He conjectured that for every disjoint system of n congruences with distinct moduli there exists an x, 1 <x < 2n, satisfying none of them. P. Erd6s [2] proved this with n2n instead of 2^ and proposed the stronger conjecture that any system of n congruence classes not covering all integers omits some x between 1 and 2 n. He proved this with 2n replaced by some constant depending only on n.

Erd6s repeated both conjectures at the number theory conferences in Boulder, Colorado [3], and Pasadena, California [4], in 1963. Prizes of $10 and $25 were announced at the former for their solution.

The first conjecture was proved by J. Selfridge [6]. In this paper we prove the second conjecture [1]. That 2n is the best possible follows from the example x_2i-1 (mod 2i), in, which covers 1, 2, .., 2n-1.

The content of our Lemma 1 was discovered independently by J. Selfridge [5], who has also proved this conjecture.

THEOREM. Let a,, a2, . *, an, bl, b2, . . , bn be given, the b's positive. Suppose there exists an integer xo satisfying none of the congruences

x-=_a, (mod b,), i =1, 2, * ,n.

Then there is such an xo among 1, 2, 3, * ,2".

LEMMA 1. Suppose the above theorem is false. Then for some n there exist congruences

x _a, (mod b,), i = 1, 2, * *,n

such that the following three conditions all hold: (A) If 1 < x < 2n, then x satisfies at least one of the congruences; but 0

satisfies none of them. (B) All the b's are prime. (C) If k of the congruences have the prime modulus p, then 2k<p.

Received by the editors March 24, 1969. Partially supported by NSF grants GP 6663 and GP 8075.

475



476 R. B. CRITTENDEN AND C. L VANDEN EYNDEN [March

PROOF. Let us assume n is the smallest positive integer for which the theorem fails. Then there exist ai and bi, 1 < i ? n, such that each x from 1 to 2n satisfies x -a, (mod b,) for some i, yet if T = {x: x a, (mod bi), 1 i? n)}, then T is nonempty. Clearly if xE T and x=x' (mod LCM(b1, b2, . , bn)), then x'ET; thus T contains negative numbers. Let xo be the greatest nonpositive element of T. Then the congruences x_ ai-xo (mod bi), i=1, 2, , n, satisfy condition (A).

Let us now suppose the congruences x--ai (mod b,), 1 ?i<n, satisfy (A). We will make a start on (B) by proving that we may assume all the moduli are prime powers. Suppose one of the congruences is x-=a (mod b), where b is not a prime power. If each prime power dividing b also divided a, then we would have b J a, in contradiction to the second condition of (A). Thus we may assume b= paq, where p is prime q> 1, ptq, and pata. Then replacing the congruence x=a (mod b) with x-a (mod pa) yields a new set of congruences for which (A) still holds. In fact, if p)a the replacement x=a (mod p) works. (The condition pta precludes 0 as a solution of the new congruence.) Continuing in this way, we see we can produce n congruences which satisfy (A), such that if x_a (mod b) is one of them, then b=pa, p prime, and p Ia if az> 1.

Assuming our n congruences are as just described, we will now show (C) must hold. Let p be a fixed prime, and suppose exactly k of our congruences have modulus p. Since 0 is a solution of no congruence, no multiple of p is a solution of any of these k congruences. Thus the multiples of p between 1 and 2n each must be a solution of at least one of the remaining n-k congruences. These all have modulus either pa with oa> 1 or else b where ptb. The solutions of the former are all multiples of p by the last paragraph. The latter we replace by the single congruence modulo pb that is equivalent to the pair of congruences x-a (mod b) and x=O (mod p), according to the Chinese remainder theorem. None of the multiples of p are lost as solutions by this replacement.

We now have n - k congruences, each of the form x -ap (mod bp), which include among their solutions p, 2p, * * * , [2n/p]p, but not 0. Then the n-k congruences x-a (mod b) have among their solutions 1, 2, . . , [2n/p], but still not 0. Recall we assumed n to be the least integer for which the theorem fails. The theorem must be true for n-k, which implies that [2n/p]<2n-k. Thus 2n/p<2n-k, or 2k<p. This is (C).

Now we return to (B). Suppose p is prime. By what has gone before we can assume we have congruences of three types

(1) x _a (mod p), where pJ(a,



1970) ARITHMETIC PROGRESSIONS 477

(2) x--a (mod pa), where a> 1 and p ja, (3) x--a (mod b), where p b.

We may assume each a is positive and less than the corresponding modulus. Since 2P-'>p for p-2, (C) implies that there exists ao, 1 : ao < p, such that x--ao (mod p) is not one of our congruences. Let Msllb, where b runs through the moduli prime to p. Choose r such that rM=ao (mod p). We claim that rM is not a solution to any of the congruences. Our choice of r eliminates the type (1) and type (2) congruences. Type (3) is out because b| rM but 0 is not a solution to any congruence. Suppose now we replace each type (2) congruence x-a (mod pa) with x_a (mod p). The integers 1, 2, * * , 2n are still all solutions of some congruence; but now so is 0. We have lost (A). The integer rM is still not a solution, however, since only multiples of p have been added. Thus condition (A) can be restored by another shift, exactly as in the beginning of the proof of this lemma. Note that we have replaced the modulus pa by p. We continue in this way until all moduli are primes. Thus (B) can be assumed.

LEMMA 2. Suppose that Si, S2, * , St are sets of integers such that Si consists exactly of ki residue classes modulo bi, i = 1, 2, * * *, t, and that (b,, b,) = 1 if i $j. Suppose n is a positive integer, and let N be the number of integers x, 1 _x _ 2 n, such that x is in none of the S's. Then if 1 s<t,we have

N > 1 + 2n(1- kb.) II (1-k,/bi)

I + ki) (I + ki). i 14)811 PROOF. For S any set, let C(S) be the characteristic function of S.

First we note that 1- I C(Si) 5 I] (1 -C(Si)), since the right side is nonnegative and the left side is nonpositive unless C(S,)

=0, i= 1, 2, . . , s, in which case both sides are 1. We see the characteristic function of the set of integers not in any S is

c( U s,) = c( n r,si) =f C(@-)Si) =(1-c(sC)) J t

=11 (1 - C(S,)) I (1 - C(S,))

i \ t 21- EC(Ss)) c (1- C(S,))

i.- i j+



478 R. B. CRITTENDEN AND C. L VANDEN EYNDEN [March

where :' indicates that at most one subscript is : s. There are ki elements of Si among any bi consecutive integers, so

2n

[2n/bi]ki E , C(S,)(r) ? [2n/bi]ki + ki. r 1

Since [2n/bi]ki.<2-ki/bi [2n/bi]ki+ki, we have 2" C(Si)(r) =2nki/bi+Ei, where JEil <ki. (Note that Ej=O if biJ2n.) More generally, the Chinese remainder theorem implies that there are kikj ... k, elements of SiCSjC O*S. among any bibj *.. b. consecutive integers, so

, C(S Cn Sin . . n So)(r) = 2nkik, ... k8/bibj ... b, + E,..jx,,

where Eij. . .zI < kik, ... k.. Then

N= E C U Si (r)

2nt

R i c(si) + Sc(si ) s()- * * * (r) rl. i==i,

= 2n - E 2nk,/bi + 1' 2nkikj/bibj- + E i=.i i,j

= 21- I kilbi) f (1- ki/bd) + E,

where

I E I = | Ei E- ,' Ej +!..| i-i

< ,: ki + ,'kik, + I + ,ki) H (I + k )1.

The lemma follows.

LEMMA 3. Suppose b, b', r, r', k, and k' are integers such that O<b s b', O k<r, O<k'_r', and b-b'+r'<r. Then there exists a positive integer u such that k+u r, k'-uO0, and

(1 - k/b)(I - k'/b') (1 - (k + u)/b)(I - (k' - u)/b').

PROOF. For u > 0 the last inequality is easily seen to be equivalent to u> b-k-b'+k'. We define u to be max (1, b-k-b'+k'), making



19701 ARITHMETIC PROGRESSIONS 479

this relation automatic. If u = 1, the first two inequalities are trivial. Otherwise k+u=b-b'+k'<b-b'+r'<r, while k'-u=b'-b+k_0.

LEMMA 4. If the theorem is false, then it fails for some n <20.

PROOF. Suppose not. Then there exists n >20 and n congruences such that the conditions of Lemma 1 hold. Suppose ki congruences have modulus Pi, PI<P2< . . * <Pt. By Lemma 2 (applied to the last t-s rather than the first s factors) we will get a contradiction if we can show

(*) 2 k,/p,) (1 - k,/pi) + k,) (1 +ki) i=a+1 / ,-' i=#+1 i$1

for some s, 1 < s _ t. We shall take s = min ([n/3 ]-1, t-1). The right side of (*) has s+1 factors. Since n= ,ki, their sum is n+s+1. The expression is maximized when all the factors are equal. Thus

( s + 1 ) ( n/3 n) (right side of (*)) s

( + 1)+ n1+ 4n/s.

Here we used that (1 +n/z)' is an increasing function of z. It can be seen from inspecting a table of primes that 7r(n - [n/3 ] + 1)

< [n/3 ] for small values of n > 20; for larger n it follows from known estimates for 7r(n). Thus if s = [n/3 ]-1, we have

t 8

ki = n - ki < n - [n/3] + 1 c (the [n/3]rd prime) < p[.1a i8+1 i=1

If we define ko= t- n/3] ki and Po=P[n/ 3], we have

(1) (left side of (*)) > 2n TI (1 -k,/p), i-SO

where ko<po, and ki [log2 pi] for i=1, 2, , s by condition (C) of Lemma 1. If s = t-1 defining po = pt and ko =k also gives (1).

For convenience, we introduce a new notation. Let m. = ki if p = pi and 0 otherwise. Then

I (1 -k,/p) = H(1-mv/p).

Since _ ki>20, the conditions ko<po and k!5 [log2 pi] for i>1 imply po? 13. In particular, m2 = 0, m3 1 l, mr6 ? 2, m7< 2, mi, < 3, and m13 5 12. If p0 = 13, then n = 20 and

TI (1-m /p)=(_1-1*/3)(1-2/5)(-/7)(1-3t/11)(1-12 /13 { )./



480 R. B. CRITTENDEN AND C. L VANDEN EYNDEN (March

In this case

2n I (1 -I - 4n3 22n13 (2n13 I (1 mV/p) - 1)

24013(210+2/3/l0Ol - 1) > O,

which implies (*). Clearly 2n/81ll(1-mp/p) > 1 is sufficient to imply (*) in general. We will show that for n> 20

II (1- Mp/p)

2 (1 - 1/3)(1 - 2/5)(1 - 2/7)(1 - 3/11) (1 - 12/13)(n-8)/12

Then

2 II (1 mp/p) > 2n13(16/1001) 13-(n-20)112

= K exp n(In 16 - In 13)/12.

We have already seen this exceeds 1 for n =20; it is clearly increasing. Thus it suffices to prove (**).

Our method will be successive application of Lemma 3 to pairs of factors of 1],(1 -mP/p). This lemma says that under cer- tain circumstances (1-m/p)(1-m'/p') may be replaced by (1-(m+u)/p)(l-(m'-u)/p') without increasing the product. Al- though Lemma 3 only guarantees a positive integer u, the operation may be repeated until m+u reaches a specified limit (namely, r) or mr'-u=O. It is easily checked that if b'2b>2, then b-b'+r'9r whether r and r' are defined by

10 r=b-1, r'=b'-1, 20 r= [log2 b] r'= [1og2 b'], or, in case b<13, 30 r= [log2 b] r'=b'-10. First we use Lemma 3 with b = 13, b' = po, k = mis, k' = ko, and r and

r' as in 10. According to Lemma 3 we can increase k and decrease k' (by the same amount) until either k+u=r=12 or k'-u=0. Since k = mis c[log2 13] = 3, we can guarantee this way that k'-u < PO10. In order to avoid a mess we redefine our m's so as to denote our new product again by fp(1 -mp/p). Now mpr ?p- 10 for p = po, m1l3 ?12, and mpn ? [log2 p] for all other p. As before, EP mp = n.

Now if mp < [log2 p] for any p < 13, we apply Lemma 3 to increase mp to equal [log2 P] by taking away from mp for p > 13. This is justi- fied by taking r and r' as in 20 if the larger prime is not po, and 30 if the larger prime is po. Since n ?20 all the primes less than 13 can be



19701 ARITHMETIC PROGRESSIONS 481

"filled up" this way. If we again redefine our m's we now have the product

II (1 - rn/p)

=(1- 113) (1 -2/5) (1- 2/7)(1 -3/11 ) II (I1- n,/p), PA 13

where mp?p-1 for p> 13. Finally we use Lemma 3 with r and r' as in 10 to stuff any remaining

mp's with p> 13 down into 13. If min gets "filled up" (hits 12) we start a new factor of the form (1-7y/13) by taking k = 0 in Lemma 3. This gives

(1-1/1/3) (1-2/5) (1- 2/7) (1-3/11) (1-12/13) l(n-8) 112] (1 -.Y./13)

where 12[(n-8)/121+y=n-8, y<12. Since it is easy to check that 1 -y/13 > (1 - 12/13)Y/112, (**) follows.

Of course it remains to show the theorem is true for n <20. This may be checked by more special arguments.

REFERENCES

1. R. B. Crittenden and C. L. Vanden Eynden, A proof of a conjecture of Erdos, Bull. Amer. Math. Soc. 75 (1969), 1326-1329.

2. P. Erdos, Remarks on number theory. IV: Extremal problems in number theory. I, Mat. Lapok. 13 (1962), 228-255. (Hungarian) MR 33 #4020.

3. , Problems 29 and 30, Proc. Conf. Number Theory (Boulder, Colorado, 1963).

4. , Extremal problems in number theory, Proc. Sympos. Pure Math., vol. 8, Amer. Math. Soc., Providence, R.I., 1965, p. 183. MR 30 #4740.

5. John Selfridge, Research announcement, Amer. Math. Soc. Annual Meeting (New Orleans, 1969).

6. , On congruences covering consecutive integers, Acta Arith. (to appear). 7. S. K. Stein, Unions of arithmetic sequences, Math. Ann. 134 (1958), 289-294.

MR 20 #17.

PENNSYLVANIA STATE UNIVERSITY, PORTLAND STATE UNIVERSITY AND

OHIO UNIVERSITY



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Any n Arithmetic Progressions Covering the First 2nIntegers Cover All Integers