THE DIOPHANTINEEQUATION x + 3"=y&#34

Internat. J. Math. & Math. Sci.

VOL. 21 NO. 3 (1998) 619-620

619

THE DIOPHANTINE

^EQUATION x + 3"=y"

S.AKHTAR ARIFandFADWA S. ABU MURIEFAH DepartmentofMathematics

GirlsCollege ofEducation AI-Riyadh,SAUDIARABIA

(ReceivedMarch 11,1996andin revised form

May

31, 1996) ABSTRACT. Theobject ofthispaperistoprove^thefollowing

THEOREM. Letrnbe odd. Thenthe diophantine equationx

+

^3"

’,

ⁿ

>_

3has onlyone solution in positive integers x,t/,rn and the unique solution isgiven by rn 5

+

6M,^x

10.33M,

7.32M

andn 3.

KEY WORDS AND PHRASES: Diphantine equation.

1992AMSSUBJECT CLASSIFICATION CODES: 11D41.

INTRODUCTION

Itis wellknown that there is nogeneral method for determiningallintegralsolutions xandV for^a given diophantine equationax

+

^bx

+

^{c dy}

^n,

^{wherea,b,cand d areintegers,a}

-

^{0,b 4ac:/:0,}

d 0, but we know that ithas onlyafinitenumberofsolutions when n

>

3 This wasfirstshown by Thue

The firstresult for thetitleequation forgeneraln isduetoLebesgue

[2]

who proved that when rn 0 thereis nosolution, forrn 1, Nagell[3]hasprovedthat it hasnosolutionandin 1993Cohn[4]

hasgiven another proof forthis case.

The proof ofthe theorem is divided into two main cases

(3, x)

¹ and

3Ix.

^It is sufficientto consider x apositive integer.

Toprovethetheorem we need thefollowing

LEMMA(Nagell[5]). Theequation3x

+

¹ /,wherenis anoddinteger E3has no solution inintegersxand for Vodd and E 1.

PROOF OFTHEOREM. Supposern 2/

+

^{1. Since theresultisknown forrn 1 weshall}

lassume that k

>

0. The casewhen x is odd, can be easilyeliminated since /’^=_0

(rood 8),

so we assume that x is even.

CASE 1: Let

(3, x)

1. First let n beodd, thenthereisno loss of generalityinconsidering n p anodd prime. Thus x

+

^3k+l /’. Thenfrom[6,Theorem 1]wehaveonly^twopossibilities and they are

where y ^a

+

^{3b and}

x+3

k=

^a+b

a=b--l(mod2)

(2)

where!/=

as

^{3b-- for}some rationalintegersaand b.

In (1)since a

+

^{3b and}F^isoddsoonly one ofaor b is oddand theotheriseven. Equating

imaginary

pans

weget _{v_t}

r--O 2r+l

620 ^$.A ARIF AND F.^SAMURIEFAH

Sob is odd Since3doesnotdividetheterminside we get b

+

3^k Hence

::[::1---(2r+1

^I9

) ap-2r-1(-32k+l)r

r=0

This isequation(1)in[6],andLemmas4and 5 in[6]showthatboththesignsareimpossible. Hence(1) givesrisetonosolutions

Nowconsiderequation(2).

By

equating imaginary partsweobtain

S.3^k

b(3a 3b).

(3)

Ifb

+

1in(3)weget

+

8.3^k 3a 3.

Thecasek 1 canbeeasily eliminated,sosupposek

>

^1then

+

8.3^k-1 a 1.

+a (25 + 3)/4

7. Hencefrom Thisequationhas theonlysolutiona

+

5, k 2andso y ₄

(2)

z s

If b ^+/-3

,

⁰

^<

^A

^<

^{k, then (3) becomes +/-}

^8.3--

^a2-

^32’,

^{and this is not possible}

modulo 3 if k A 1

>

0. So k A 1 0, that is

+

8 a

32(-I),

and we canreject the positive sign modulo3. Sowehavea 32(k-l) _8,whichhastheonlysolutiona

+

1, k 2

ani

^{z 10 Finallyifb -4-3 then}

⁺

^{8 3a2}

^32+I,

and this isnottruemodulo3.

Nowifniseven, thenfromtheaboveit is sufficienttoconsidern 4,hence

(y2 _+x) (V2 z) 32

⁺

Since

(3, x)

I,weget

y2+x=32+

and

y2_x=l,

by addingthesetwoequationsweget

2V

²

32+I +

^{1,which is}impossible modulo3.

CASE2. Let

3[z.

Thenofcourse

31!/.

Supposethat x 3X,y 3Ywhere u

>

0,v

>

0and

(3,X) (3,Y)

^{1 Then}

32X + ^32k+1

3"Y" There are threepossibilities.

2u min(2u,2k

+

1,

nv).

^Then by cancelling

32u

^we get X

+

^32(k-u)+l

3"w-2"Y

’, and considering this equation modulo 3 we deduce that nv-2u-O, then

z2+

^32(-u)+I

yn,

with

(3,X)

^{1. If k-u 0, this} equation hasno solution[3,4] ^{andif k-u}

>

0, as proved abovethis equation has a solutiononly ifk u 2 andn 3, sonv 3v 2u that is

3[u,

letu 3M then k 2

+

^{3Mand m 5}

+

^{6M. Sothisequation} has a solutiononlyif m 5

+

6M and the solutionis given by

X

10,

Y

7. Hence the solution of our title equation is x 10.3 10.3^TM and y 7.3 7.3

TM.

2 2k

+

¹ min(2u,^2k

+ 1,nv)

^Then

32u-2-X

²

+

¹

3"v-2-Y

and considering this equation modulo3 wegetnv 2 1 0,sonisoddand

3(T ’-- X) +

¹

Y",

by thelemma this cquationhas nosolution.

3. nv

min(2u,

2k

+ ^1,nv).

^Then

^32-"X

²

+ ^{32+-" Y’}

^{andthis ispossiblemodulo3only}

if 2u nv 0or 2k

+

^{1 nv} 0 and both ofthesecaseshavealready beendiscussed Thisconcludes theproof.

REFERENCES

THUE, A.,

^UberdieUnl6sbarkeitderGleichunga:r

+

^bz

+

^{c dy in}grossen ganzen Zahlen^z and

,

^Arch.Math.ogNatgrvidenskab,Kristiania, BdXXXIV (l 916),I-6.

[2] LEBESGUE, V.A.,

Sur rimpossibilite ^ennombresentiresderlequationx

!/2 +

^1,Nouvelles

Ann.

desMath.9

(I) (1850),

^175-I$I.

[3]

^NAGELL,

T.,

Sur rl impossibilite’ ^de quelques

&luations

deux indermindes, Norsk Math.

ForeningsSkrifler,Kristiania,Set.

I,

13(1923),65-82.

[4]

^COHN,

J.H.E,

Thediophantineequationx²

+

³ y’, Glasgo^Math.

J.,

35

(1993),

^203-206.

[5] NAGELL,

T.,

Contributions ^tothe theory ofacategory of diophantine equations of^the second degreewith twounknown,Nova Acta

Reg.

Soc. Upsal, Set

IV,

¹⁶

(1955),

^1-35.

[6] COHN, I.H.E,

^Thediophantine^equation

2 +

^c

^V",

^ActaAritb.65

(4) (1993),

^367-381