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Article 03.3.5

Journal of Integer Sequences, Vol. 6 (2003), 2

3 6 1 47

On Obl´

ath’s problem

Alexandru Gica and Laurent¸iu Panaitopol

Department of Mathematics

University of Bucharest

Str. Academiei 14

RO–70109 Bucharest 1

Romania

alex@al.math.unibuc.ro

pan@al.math.unibuc.ro

Abstract. _{In this paper we determine those squares whose decimal representation consists}

of k_≥2 digits such that k₋1 of them equal.

1

Introduction

R. Obl´ath [5] succeeded in almost entirely solving the problem of finding all the numbers

nm _{(n, m ∈ N, n ≥ 2, m ≥ 2) that have equal digits. The special case m = 2 is a very} well known result, although its proof involves no difficulty. In this connection, the following question naturally arises: is it possible to determine all of the squares having all digits but one equal?

The answer is given by

Theorem 1.1 _{The squares whose decimal representation makes use of k ≥ 2 digits, such}

that k₋1 of these digits are equal, are precisely 16, 25, 36, 49, 64, 81, 121, 144, 225, 441, 484, 676, 1444, 44944, 102i_{, 4·10}2i _{and 9·10}2i _{with i≥1.}

When we are looking for the squares with k digits among whichk₋1 digits equal 0, we immediately get that the corresponding numbers are 102i_{, 4·10}2i _{and 9·10}2i _{with i≥1.}

A simple computation shows that the numbers with at most 4 digits verifying the condi-tion in the statement are just the ones listed above.

Since every natural number can be written in the form 50000k_±r with 0_≤r _≤ 25000, and (50000k_±r)2 _{≡ r}2 _{(mod 100000), we compute r}2 _{for r ≤25000 and find that the last}

We select the squares such that 4 of the last 5 digits are equal, because these point out the possible squares with k _≥ 5 digits, k ₋1 digits of them being equal. If one excludes the numbers for which there are k₋1 digits equal to 0, then there still remain 22 types of numbers, namely:

a1 = 1_{· · ·}121 a7 = 4_{· · ·}441 a13 = 4_{· · ·}4944 a18 = 7_{· · ·}76

a2 = 1_{· · ·}161 a8 = 4_{· · ·}449 a14 = 4_{· · ·}45444 a19 = 8_{· · ·}81

a3 = 2_{· · ·}224 a9 = 4_{· · ·}464 a15 = 4_{· · ·}49444 a20 = 8_{· · ·}89

a4 = 2_{· · ·}225 a10 = 4_{· · ·}484 a16 = 5_{· · ·}56 a21 = 9_{· · ·}929

a5 = 4_{· · ·}41444 a11 = 4_{· · ·}4544 a17 = 6_{· · ·}656 a22 = 9_{· · ·}969

a6 = 4_{· · ·}4144 a12 = 4_{· · ·}4644

One will show that, among these numbers with k_≥5 digits, only 44944 is a square. The exclusion of the other numbers can be carried out fairly easily in certain cases, as we show in_§2. In the other cases we will solve equations of the type

x2₋dy2 =k (1)

(where d, k _∈Z∗_{, d > 0 and} √_{d 6∈}Z_{) in integers. The literature concerning equation (1) is}

rather extensive. In this connection, we mention [1, 2, 3, 4].

We now recall the solving method (in accordance with [2]). We denote by (r, s) the minimal positive solution to the equation

x2₋dy2 = 1 (2)

and by ε = r+s√d. We determine the “small” solutions to equation (1) (if any). They generate all the solutions.

Theorem 1.2_{We denote by µi =ai+bi}√_{d, i= 1, m all the numbers with the property that}

(ai, bi)is a solution in nonnegative integers to equation (1) withai ≤

p

|k_|εandbi ≤

p

ε_|k_|/d

(if any). If x and y are solutions to (1) then there exist i, n _∈ Z _{such that 1 ≤i ≤ m and}

x+y√d=_±µiεn or x+y √

d=_±µ¯iεn.

We will use this theorem in _§3.

2

Excluding the simple cases

We assume in this section thatk_≥5 and an is a square, hence 9an is a square as well. Make use of simple reasonings, we shall show that this fact is impossible. To this end, we use the symbol of Legendre in some cases.

The 16 cases which have to be excluded will be exposed in a concise form, inasmuch as some of them are quite similar:

a3, a4, a20; a2, a15, a18; a11, a17.

1. _{We have 9a2 = 10}k_{+ 449≡(−1)}k_{+ 9 (mod 11). But} ¡8

11

¢

=¡10₁₁¢=₋1.

2. _{It follows by 9a3 = 2(10}k_{+ 8) = (4x)}2 _{that 2}k−3₅k _{= (x−1)(x+ 1). Since (x−}

1, x+ 1) = 2, we have 5k _{| x+ε with ε ∈ {−1,1}, whence x+ 1 ≥ 5}k_{. Consequently} 2k−3_·5k _=x2 _−1≥5k₍₅k_{−2), hence 2}k−3 _≥5k_−2.

3. _{By 9a4 = 2 · 10}k _{+ 25 = (5x)}2_{, we have 2}k+1 _{· 5}k−2 _{= (x− 1)(x + 1), whence}

2k+1_·5k−2 _≥5k−2₍₅k−2_−2).

4. _{We have 9a5 = 4·10}k_{− 27004 = (2x)}2_{, whence 10}k _{−6751 = x}2_{. When k is an}

odd number, we have 10k _{−6751 ≡ 2 (mod 11), but} ¡2 11

¢

= ₋1. If k = 2h with h _≥ 3, then (10h _−x)(10h _{+x) = 6751, whence 10}h _{−x = a, 10}h _{+x = b, where we have either} (a, b) = (1,6751) or (a, b) = (43,157). Since 2_·10h _{=a+b, it follows that either 2·10}h _{= 6752} or 2_·10h _{= 200, althoughh≥3.}

5. _{By 9a6 = 4·10}k_{−2704 = (4x)}2 _{it follows that 5}2_·10k−2_{−169 =x}2_{. Since k≥5, we}

get that x2 _{= 5}2_·10k−2_{−169≡ −}4 _169≡4 _3.

6. _{We havea9 = 4·11· · ·16, but 11· · ·16 does not occur among the numbers ai.} 7. _{We havea10= 4a1, and we shall get the contradiction after we studya1 for k≥5.}

8. _{We have 9a11 = 4·10}k_{+ 896 = (8x)}2_{, hence 2}k−4₅k_{+ 14 = x}2_{. It follows that x}...2.

Thereforex2...4, and k = 5. In this case we getx2 = 6264.

9. _{We havea12= 4a2, but a2 6=x}2_.

10. _{By 9a15 = 4·10}k_{+44996 = (2x)}2 _{it follows that 10}k_{+11249 =x}2_{. Then 10}k_{+11249 ≡}

(₋1)k_{+ 7 (mod 11), but} ¡6 11

¢

=¡ 8 11

¢

=₋1.

11. _{Since 4a16= 22· · ·2}

| {z }

k

4 and a3 ₆=x2_{, it follows thata166=y}2_.

12. _{We have 9a17 = 6· 10}k _{−96 = (4x)}2_{, hence 3 ·2}k−3₅k _{−6 = x}2_{. But x}2...4 and

3_·2k−3₅k...4 (because _{k ≥5).}

13. _{We have 9a18 = 7·10}k_{−16≡7(−1)}k_{−5 (mod 11). But} ¡2

11

¢

=¡10 11

¢

=₋1.

14. _{By 9a20 = 8· 10}k _{+ 1 = x}2 _{it follows that (x −1)(x + 1) = 2}k+3₅k _{and then}

2k+3₅k _≥5k₍₅k_−2).

15. _{We havea21≡2 (mod 9).} 16. _{We havea22≡6 (mod 9).}

3

The six difficult cases

Just as in the previous cases, the numbers under consideration havek _≥5 digits, and k₋1 of these digits are equal.

1. _{For a1 = 11· · ·121 =x}2 _{it follows that (10}k_{−1)/9 + 10 =x}2_{. We denote y= 3x and,}

since k_≥5, we have y >316 and

For k = 2m we have (10m_−y)(10m _{+y) = −89, whence we get both 10}m_{−y = −1 and} 10m_{+y= 89, which is a contradiction.}

For k= 2m+ 1, we denotez = 10m _{and then}

y2₋10z2 = 89.

The primitive solution of the Pell equation

x2₋10y2 = 1

is (r, s) = (19,6). Making use of Theorem 1.2 in the Introduction, we getbi ≤

q

89 10

¡

19 + 6√10¢, hence bi ≤18. We find b1 = 8, a1 = 27, and b2 = 10, a2 = 33.

It follows that either

y+z√10 = ³_±27_±8√10´ ³19 + 6√10´t

or

y+z√10 =³_±33_±10√10´ ³19 + 6√10´t,

with t_∈Z_{. Sincey >0, z >0, we have only the solutions}

y+z√10 = ³27_±8√10´ ³19 + 6√10´t

and

y+z√10 =³33_±10√10´ ³19 + 6√10´t,

with t _∈Z_{. Since} 27+8√10

19+6√10 <2 and

33+10√10

19+6√10 <2, it follows that t ∈ N. Let

¡

19 + 6√10¢t =

at+bt √

10, at, bt∈N, a0 = 1, b0 = 0. Fort ≥1, we have the following equalities:

at= 19t+Ct219t−2·62·10 +· · · (4)

and

bt=Ct119t−1·6 +Ct319t−3·63·10 +· · · . (5) We have at ≡ 1 (mod 3) and bt ≡ 0 (mod 3). Since z = 10m ≡ 1 (mod 3), we only have one of the situations

y+z√10 =³27₋8√10´ ³19 + 6√10´, t _∈N_{, (6)}

and

y+z√10 =³33 + 10√10´ ³19 + 6√10´, t_∈N_{. (7)}

a) In the case of the relation (6), we have the identity

z = 10m = 27bt−8at. (8)

For m= 2, the equation (3) takes the form y2₋₁₀5 _{= 89, and has no integer solutions.}

For m_≥3, it follows that 8_|bt. By (5) we have bt≡6t·19t−1 (mod 8), whence t= 4h. It then follows by (4) and (5) that

at ≡64h·102h (mod 19) and 19|bt. The equation (3) takes the form

104f+1+ 89 =y2. (9)

Consequently m= 2f, and we get (9) again, which is a contradiction.

Just as in the previous case, we make use of Theorem 1.2 and, fory >0,z >0, we get that either

y+z√10 =³3 + 2√10´ ³19 + 6√10´t (12)

or

y+z√10 =³₋3 + 2√10´ ³19 + 6√10´t, (13)

with t_∈N_.

a) By (12) we obtain the equation:

2_·10m = 3bt+ 2at. (14)

Since m_≥ 2, it follows by (5), (6) and (14) that 3_·6t_·19t−1_{+ 2·19}t _{≡0 (mod 8), that is,}

9t+ 19 _≡ 0 (mod 4), whence t = 4h+ 1. By (4) and (5) we get that z _≡ 3_·64h+1 _·102h (mod 19), that is, 10m _{≡ 3}2 _·64h_·102h _{(mod 19), whence} ¡10m

19

¢

= 1. Consequently m is an even number.

On the other hand, it follows by (14) that 3bt + 2at ≡ 0 (mod 5), that is, at ≡ bt (mod 5). By (4) and (5) it follows that at ≡ (−1)t (mod 5) and bt ≡ (−1)t−1t (mod 5), whence t _≡ 4 (mod 5). We have (3 +√10)5 _{= 4443 + 1405}√_{10 ≡ −53 (mod 281), hence}

(19 + 6√10)5 ₌¡_{(3 +}√₁₀₎5¢2 _≡532 _{≡ −1 (mod 281). Consequently}

y+ 2_·10m√10 =³3 + 2√10´ ³19 + 6√10´t+1³19₋6√10´

=³₋63 + 20√10´ ·³19 + 6√10´5

¸(t+1)/5

≡ −63 + 20√10 (mod 281).

We have taken into account that t_≡4 (mod 5) and t is an odd number.

We have 2 _· 10m _{≡ 20 (mod 281), that is, 10}m−1 _{≡ 1 (mod 281). Since 10}7 _{≡ 53}

(mod 281), it follows that 1014 _{≡ −1 (mod 281) and 10}28 _{≡ 1 (mod 281). Thus we have}

ord 10 = 28 inZ_{281, whence 28|m−1, which is a contradiction since m is even.}

b) It follows by (13) that

2_·10m =₋3bt+ 2at. (15)

Sincem _≥2, it follows that ₋3bt+ 2at ≡0 (mod 8). We get by (4) and (5) that t= 4h+ 3 and then 2_·10m _{≡ −3·6}4h+3_·102h+1 _{(mod 19). Therefore} ¡10m

19

¢

= 1, and m is even. Also by (15) we get at+bt≡0 (mod 5), and in view of (4) and (5) we have t≡1 (mod 5). The relation (13) can be written as:

y+ 2_·10m√10 =³₋3 + 2√10´ ³19 + 6√10´t−1³19 + 6√10´

=³63 + 20√10´ µ³19 + 6√10´5

¶(t−1)/5

≡63 + 20√10 (mod 281).

Just as in the case a), it follows that 2_·10m _{≡ 20 (mod 281). The relation 10}m−1 _{≡ 1}

3. _{Fora8 = 44· · ·49 = x}2 _{we have 4·}10k−1

wheret is a natural number. a) It follows by (16) that

2_·10m = 9bt−2at.

Since 2_·10m _{≡ 2 (mod 3), and on the other hand we have by (4) that a}

t ≡ 1 (mod 3), we get the contradiction 2_{≡ −}2 (mod 3).

b) It follows by (17) that

2_·10m = 2at+ 9bt. (18)

Then bt≡2at (mod 5), whence t≡3 (mod 5). We also have bt+ 2at≡0 (mod 4), hence t is odd.

The equality (17) takes the form

whence either

y+z√10 =³21 + 4√10´ ³19 + 6√10´t, t_∈N_{, (20)}

or

y+z√10 =³21₋4√10´ ³19 + 6√10´t, t_∈N∗_{. (21)}

a) By (20) it follows that

5_·10m = 21bt+ 4at,

For m = 0 and m = 2, equation (19) has no integer solutions, hence we may consider

m _≥ 3. We have by (22) that bs ≡ 2as (mod 8). Hence it follows by (4) and (5) that

and none of these systems has solutions.

If k= 2m+ 1, then m _≥2. With z = 10m _{we have the equation:}

The initial solutions (a, b) of the equation are (57,10), (147,44), (153,46), hence the solutions with y >0,z >0 are given by the identities:

y+ 10m√10 =³57₋10√10´ ³19 + 6√10´t, (24)

y+ 10m√_{10 =}³_{57 + 10}√₁₀´ ³_{19 + 6}√₁₀´t_{, (25)}

y+ 10m√10 =³147₋44√10´ ³19 + 6√10´t, (26)

y+ 10m√10 =³147 + 44√10´ ³19 + 6√10´t, (27)

y+ 10m√10 =³153₋46√10´ ³19 + 6√10´t, (28)

y+ 10m√10 =³153 + 46√10´ ³19 + 6√10´t, (29)

where t _∈ N_{. We get by (24) that 10}m _{= 57bt−10at, hence 10}m _{≡ −at (mod 3), which}

yields the contradiction 1_{≡ −}1 (mod 3). If (27) was true, then 10m _{= 147b}

t+ 44at. Since at ≡1 (mod 3) and 10m ≡1 (mod 3), the contradiction 1_≡44 (mod 3) follows.

In the case when (28) holds, we get 10m _{= 153b}

t−46at, whence the contradiction 1≡ −46 (mod 3).

We still have to study three situations. a) We have by (25) that

10m = 57bt+ 10at. (30)

Since m _≥ 2, it follows that bt + 2at ≡ 0 (mod 4). By (4) and (5) we have 6t(−1)t−1 + 2(₋1)t_{≡0 (mod 4). Therefore 3t−1≡0 (mod 2), and we get thattis odd. It then follows} that 19_|at, and by (30) we deduce the contradiction 19|10m.

b) It follows by (26) that

10m = 147bt−44at. (31)

Just as in the case a), by considering congruences (mod 4), we get thatt is even.

Also by (31) we have 2bt+at ≡0 (mod 5), whence by (4) and (5) we gett≡3 (mod 5). Then (t+ 2)/5 is an even natural number. The relation (26) takes the form

y+ 10m√10 = ³147₋44√10´ ³19 + 6√10´t+2³19₋6√10´2

≡³147₋44√10´ ³721₋228√10´ (mod 281)

≡³147₋44√10´ ³₋122 + 53√10´ (mod 281).

It then follows that 10m _{≡13159 (mod 281) ≡ −48 (mod 281), hence} ¡10m

281

¢

=¡−48

281

¢

. One directly gets a contradiction, observing that ¡₂₈₁10¢=¡−1

281

¢

=¡₂₈₁16¢= 1 and ¡₂₈₁3 ¢=₋1. c) By (29) we have the equation:

Form = 2, we get the number 102249 which is not a square. Hencem_≥3.

It then follows fory, z >0 that either

Fory, z >0 we have either

y+z√5 =³3 + 4√5´ ³9 + 4√5´t, (37)

or

y+z√5 =³₋3 + 4√5´ ³9 + 4√5´t, (38)

wheret is a natural number. We put ¡9 + 4√5¢t=et+ft

√

5 and then

et = 9t+Ct2·9t−2·42·5 +· · · , (39)

ft = 4t·9t−1+Ct3·9t−3 ·43·5 +· · · (40)

It follows by (37) and (38) that z = 4_·10m _{= 4e}

t±3ft, hence 4et±3ft ≡ 0 (mod 8). By (39) and (40) we get 4_±4t _≡0 (mod 8), whence t= 2h+ 1.

By 4_·10m _{= 4e}

t±3ft we infer 4·10m ≡ et (mod 3). Since t is odd, we have et ≡ 0 (mod 3), and the contradiction 4_·10m _{≡0 (mod 3).}

Remarks. It would be interesting to solve the similar problem involving numbers written with respect to some basisb _≥2.

It might be more difficult to consider the same problem imposing the condition all-but-one-equal-digits to higher powers, instead of squares.

References

[1] G. Chrystal Algebra. An Elementary Textbook. Part II, Dover, New York, 1961, pp. 478–486.

[2] A. Gica, Algorithms for the equationx2_−dy2 _{=k.Bull. Math. Soc. Sci. Math. Roumanie}

38₍86_{) (1994-1995), 153–156.}

[3] R.E. Mollin,Fundamental Number Theory with Applications. C.R.C. Press, Boca Raton, 1998, pp. 299–302, 232.

[4] I. Niven, H.S. Zuckerman, H.L. Montgomery,An Introduction to the Theory of Numbers. Fifth edition. John Wiley & Sons, Inc., New York, 1991, pp. 346–358.

[5] R. Obl´ath, Une propriet´e des puissances parfaites. Mathesis 65 _{(1956), 356–364.}

2000 Mathematics Subject Classification: Primary 11A63; Secondary 11D09, 11D61. Keywords: square, equal digits, Pell equation

Received March 12 2003; revised version received September 15 2003. Published in Journal of Integer Sequences, October 2 2003.