Handling a large bound for a problem on the generalized Pillai equation ±r a x ±s b y =c
aa r X i v : . [ m a t h . N T ] D ec Handling a large bound for a problem on the generalized Pillai equation ± ra x ± sb y = c . Reese ScottRobert Styer (correspondence author), Dept. of Mathematical Sciences, Villanova University, 800 LancasterAvenue, Villanova, PA 19085–1699, phone 610–519–4845, fax 610–519–6928, [email protected] 19 Dec 2011
Abstract
We consider N , the number of solutions ( x, y, u, v ) to the equation ( − u ra x + ( − v sb y = c innonnegative integers x, y and integers u, v ∈ { , } , for given integers a > b > c > r > s >
0. Previous work showed that there are nine essentially distinct ( a, b, c, r, s ) for which N ≥
4, exceptpossibly for cases in which the solutions have r , a , x , s , b , and y each bounded by 8 · or 2 · . Inthis paper we show that there are no further cases with N ≥ N = 3for an infinite number of ( a, b, c, r, s ), even if we eliminate from consideration cases which are directlyderived from other cases in one of several completely designated ways. Our work differs from previouswork in that we allow x and y to be zero and also allow choices of ( u, v ) other than (0 , The problem of finding N , the number of solutions ( x, y, u, v ) to the equation( − u ra x + ( − v sb y = c (1)in nonnegative integers x, y and integers u, v ∈ { , } , for given integers a > b > c > r > s > N > a, b, r, s, x, y ) < · (or, in some cases, 2 · ). Thepurpose of this paper is to show that there are exactly nine essentially different cases with N > set of solutions to (1) which we write as( a, b, c, r, s ; x , y , x , y , . . . , x N , y N )and by which we mean the set of solutions ( x , y ), ( x , y ), . . . , ( x N , y N ) to (1), with N >
2, for givenintegers a , b , c , r , and s . We say that two sets of solutions ( a, b, c, r, s ; x , y , x , y , . . . , x N , y N ) and( A, B, C, R, S ; X , Y , X , Y , . . . , X N , Y N ) belong to the same family if a and A are both powers of thesame integer, b and B are both powers of the same integer, there exists a positive rational number k suchthat kc = C , and for every i there exists a j such that kra x i = RA X j and ksb y i = SB Y j , 1 ≤ i, j ≤ N .One can show [14] that each family contains a unique member ( a, b, c, r, s ; x , y , x , y , . . . , x N , y N ) with thefollowing properties: gcd( r, sb ) = gcd( s, ra ) = 1; min( x , x , . . . , x N ) = min( y , y , . . . , y N ) = 0; and neither a nor b is a perfect power. We say that a set of solutions with these properties is in basic form .1he associate of a set of solutions ( a, b, c, r, s ; x , y , x , y , . . . , x N , y N ) is the set of solutions ( b, a, c, s, r ; y , x , y , x , . . . , y N , x N ).A subset of a set of solutions ( a, b, c, r, s ; x , y , x , y , . . . , x N , y N ) is a set of solutions with the same( a, b, c, r, s ) and all it pairs ( x, y ) among the pairs ( x i , y i ), 1 ≤ i ≤ N . Note that this subset may be (and, inour usage, usually is) the set of solutions ( a, b, c, r, s ; x , y , x , y , . . . , x N , y N ) itself.We are now ready to state the result of this paper: Theorem 1.
Any set of solutions ( a, b, c, r, s ; x , y , x , y , . . . , x N , y N ) to (1) with N > must be in thesame family as a subset (or an associate of a subset) of one of the following: (3 , , , ,
2; 0 , , , , , , , , , , ,
2; 0 , , , , , , , , , , , , ,
2; 0 , , , , , , , , , , ,
2; 0 , , , , , , , , , , , , ,
1; 0 , , , , , , , , , , ,
2; 0 , , , , , , , , , , ,
7; 0 , , , , , , , , , , ,
1; 0 , , , , , , , , , , ,
1; 0 , , , , , , , N = 3, even if we consider only sets of solutions inbasic form (see (11) through (18) in Section 2). If, for a given choice of ( a, b, c, r, s ), (1) has two solutions ( x i , y i ) and ( x j , y j ), 1 ≤ i, j ≤ N , we have ra min ( x i ,x j ) ( a | x j − x i | + ( − γ ) = sb min( y i ,y j ) ( b | y j − y i | + ( − δ ) (2)where γ, δ ∈ { , } .In the following lemma, we summarize some results which follow from the proof of Theorem 2 of [14]. Lemma 1.
Any set of solutions violating Theorem 1 above must be in the same family as a basic form (orthe associate of a basic form) which satisfies one of the following: ( ra, sb ) = 1 , max( a, b, r, s, x, y ) < · , x < x < x < · · · < x N , y < y < y < · · · < y N , (3)( ra, sb ) = 1 , max( a, b, r, s, x, y ) < · , x < x < x < · · · < x N , y < y < y < · · · < y N , (4)( ra, sb ) = 1 , max( a, r, s, x, y ) < · , x < x < x < · · · < x N , y = y < y < · · · < y N , (5)We will also need the following two lemmas from [14]. Lemma 2. (Lemma 13 of [14]) Suppose ( ra, sb ) = 1 and suppose (1) has four solutions ( x , y ) , ( x , y ) , ( x , y ) , ( x , y ) with x < x < x < x . Let Z = max( x , y , y , y , y ) . Then a x − x ≤ Z, s ≤ Z + 1 . emma 3. (Lemma 17 of [14]) Let a > and b > be relatively prime integers. For ≤ i ≤ m , let p i beone of the m distinct prime divisors of a . Let p g i i || b n i ± , where n i is the least positive integer for whichthere exists a positive integer k such that | b n i − kp i | = 1 , and ± is read as the sign that maximizes g i . Write σ = X i g i log( p i ) / log( a ) . Then, if a x | b y ± , where the ± sign is independent of the above, we must have a x | a σ y. Define σ a ( b ) to be the σ of this lemma, and let σ b ( a ) be the σ of this lemma with the roles of a and b reversed.In the course of this paper, we will often need to show that a given set of three solutions does not have afourth solution. We eliminate the possibility of a fourth solution by one of three methods: using the methodknown as ‘bootstrapping’ (see [4] or [18]), using bounds derived from LLL basis reduction (see [17]), or using x to calculate y and seeing if y is an integer (or using y to calculate x ).The technique known as ‘bootstrapping’ assumes one knows the values of a , b , r , s , x , and y . Forsimplicity of exposition assume γ = δ = 1 (the other cases are only slightly more complicated) and consider(2) with ( i, j ) = (3 , ra x ( a x − x −
1) = sb y ( b y − y − . Let ord( n, p ) be the least positive integer such that p | n ord( n,p ) −
1. Since ( ra, sb ) = 1, for each prime p | sb y , we have ord( a, p ) | x − x . Let x = lcm { ord( a, p ) : p | sb y } so x | x − x . Similarly, define y = lcm { ord( b, p ) : p | ra x } so y | y − y . Now we begin the bootstrapping steps. For each prime p | a x − p sb , ord( b, p ) must divide y − y ; setting y = lcm( y , { ord( b, p ) : p | a x − } ),we have this new y | y − y . For each prime p | b y − p ra , ord( a, p ) | x − x ; set x = lcm( x , { ord( a, p ) : p | b y − } ), so this new x | x − x . We alternately use x to find a larger y ifpossible, the new y to find a larger x if possible, etc., continuing to bootstrap back and forth until x or y exceeds 8 · , in which case x − x or y − y must exceed this bound, contradicting Lemma 1, so nofourth solution exists.For the LLL basis reduction algorithm, we follow the exposition in [17]. We have ± c = ra x − sb y so ra x sb y = 1 ± csb y and thus log (cid:16) rs (cid:17) + x log( a ) − y log( b ) = log (cid:16) ± csb y (cid:17) . (6)Since | log(1 ± x ) | < x for 0 < x < .
5, one derives from (6) that, if c/ ( sb y ) < . (cid:16) rs (cid:17) + x log( a ) − y log( b ) < csb y so log (cid:16) rs (cid:17) + x log( a ) − y log( b ) < cs e − log( b ) y . C = 10 , we set A = (cid:18) C log( a )]0 [ − C log( b )] (cid:19) , Y = (cid:0) (cid:2) − C log (cid:0) rs (cid:1)(cid:3) (cid:1) , where we set [ X ] to be the closest integer to X . Note that rows in the computer algebra program Maplecorrespond to columns in [17]. Let B be the LLL basis reduction of the rows of A , and let b and b be therows of B . Set b ∗ = b − b · b b · b b . Further, define the vector σ = Y B − , define the number σ = σ [2], and let { σ } = σ − [ σ ] be the distance from σ to the nearest integer. In our context, Lemma VI.1 in [17] becomes Lemma 4.
Let S = (cid:0) · (cid:1) , T = 8 · + 0 . , c = max(1 , || b || / || b ∗ || ) , c = rs , c = log( b ) , and c = c − { σ }|| b || . Assume c/ ( sb y ) < . . If c > S + T then y ≤ c (cid:18) log( Cc ) − log (cid:18)q c − S − T (cid:19)(cid:19) . (7)Given a , b , c , r , s , and verifying that c/ ( sb y ) < .
5, we can often use this lemma to find that max( x , y ) < min( a, s − a , b , r , s , and either a potential x or a potential y . Suppose we have a bound c < and suppose y ≥ b ≥ b y > . Using | log(1 ± x ) | < x for x < .
5, we see that (6) implies (cid:12)(cid:12)(cid:12) log (cid:16) rs (cid:17) + x log( a ) − y log( b ) (cid:12)(cid:12)(cid:12) < − . So we have to more than 1000 places of accuracy, y = x log( a )log( b ) + log( r/s )log( b ) (8)and x = y log( b )log( a ) + log( s/r )log( a ) . (9)If we know a , b , r , s , and x , we can calculate y from (8) and if this y is not an integer to 1000 places ofaccuracy, then no fourth solution can exist. Similarly, given y we can calculate x from (9) and hope that x is not an integer, thus showing no fourth solution can exist.We will also need the following. Lemma 5.
Suppose (1) has three solutions ( x , y , u , v ) , ( x , y , u , v ) , and ( x , y , u , v ) and furtherassume that the following four conditions hold:1.) x < x < x and y < y < y ,2.) u = v ,3.) any solution ( x, y ) to (1) such that x > x and y > y must also satisfy x ≥ x and y ≥ y ,4.) R = ra x gcd( ra x ,sb y ) > and S = sb y gcd( ra x ,sb y ) > .Then x − x | x − x and y − y | y − y .Proof. Suppose we have three solutions to (1) satisfying all four conditions of the lemma. Considering (2)with ( i, j ) = (1 ,
2) and (1 ,
3) we have R ( a x − x + ( − γ ) = S ( b y − y + ( − δ )4nd R ( a x − x + ( − γ ) = S ( b y − y + ( − δ ) . Since u = v , we must have γ = δ and γ = δ . Let α = 1 if γ = γ = 1, otherwise let α = 0. Let t = a x − x + ( − γ S = b y − y + ( − γ R and T = a x − x + ( − γ S = b y − y + ( − γ R .
Note that t and T are both integers.Let g = gcd( x − x , x − x ) and g = gcd( y − y , y − y ). Let k be the least integer such that b k + ( − α is divisible by R . Then k must divide both y − y and y − y , so that k divides g , and b g + ( − α = Rl for some integer l . (Note that, when α = 0, 2 n || k implies 2 n || y − y when γ = 0 and 2 n +1 | y − y when γ = 1, similarly for y − y , so that, since min( γ , γ ) = 0, we have 2 n || g .) Similarly, a g + ( − α = Sl for some integer l . Since g divides both x − x and x − x , l divides t and T . There must be an integer j which is the least positive integer such that b j + ( − α is divisible by Rl , and j must divide both y − y and y − y , so that j divides g . Therefore, l | l .A similar argument with the roles of a and b reversed shows that l | l , so that l = l , and we have ra x ( a g + ( − α ) = sb y ( b g + ( − α ) . (10)(10) shows that ( x + g , y + g ) is a solution to (1). If x + g = x , then, using Condition 3 in theformulation of the lemma, we see that we must have x + g > x , which is impossible by the definition of g . So x + g = x and, similarly, y + g = y .When N = 3, we find many sets of solutions. Here we list several types of sets of solutions, each one ofwhich generates an infinite number of basic forms (and therefore an infinite number of families) giving threesolutions to (1). We list these sets of solutions in the form ( a, b, c, r, s ; x , y , x , y , x , y ):( a, a kd + ( − u + v a d + ( − u , a d b + ( − u + v +1 h , b + ( − v h , a d + ( − u h ; 0 , , d, , kd,
2) (11)where a and b = a kd +( − u + v a d +( − u are integers greater than 1, d and k are positive integers, h = gcd( a d +( − u , b +( − v ), and u and v are in the set { , } . When u = 0, we take k − v odd; when ( u, v ) = (1 , a d ≤
3. When a = d = 2 and ( u, v ) = (1 , k to be a half integer. When k = 2 and u − v isodd, the same choice of ( a, b, r, s ) as in (11) gives the additional set of solutions( a, a d + ( − v , a d + ( − v h , a d + ( − v h , a d + ( − v +1 h ; 0 , , d, , d, . (12)Other sets of solutions can be constructed with specified values of a . When a = 3 we have(3 , g + ( − v , g +1 + ( − v v − α , g − + ( − v )2 v − α , − v + α ; 0 , , , , g,
3) (13)5here v ∈ { , } , g is a positive integer, α = 0 when 2 | g − v , α = 1 when g is odd and v = 0, and α = 2when g is even and v = 1.When a = 2, we have (2 , g + ( − v , g + ( − v +1 , ,
1; 0 , , g − , , g,
1) (14)where v ∈ { , } and g is a positive integer.Also, it is easy to construct sets of solutions for which x = y = y = 0. For example, we have, for a even and x >
0, ( a, a x ± , a x ± , , a x ∓
1; 0 , , x, , x, . (15)More generally, ( a, b, a x + ( − t m , − m , a x + ( − t +1 m ; 0 , , x , , x , y ) , (16)with b y = a x +( − t + w +1 a x +( − w +1 a x +( − t +1 , where x > x > x | x , and a x ≡ ( − w mod a x +( − t +1 m , for t ∈ { , } , w ∈ { , } , and m = 1 or 0 according as a is odd or even.We also find an infinite family for which gcd( a, b ) > a, b, c, r, s ; x , y , x , y , x , y ) = ( a, ta, a ( t + ( − u + v +1 ) h , ta + ( − v h , a + ( − u h ; 0 , , , , m + 1 ,
2) (17)where m ≥ t = a m +( − v a +( − u is an integer, h = gcd( ta + ( − v , a + ( − u ), and u and v are inthe set { , } . Closely related to (17) is the following:(2 , t, t + 4 h , t + 1 h , h ; 0 , , , , m + 2 ,
2) (18)where m ≥ − t = m +13 , h = 3 or 1 according as m ≡ u, v ∈ { , } .The Maple TM worksheets with the calculations for the following sections can be found at [19]. x < x < x < x , y < y < y < y Lemma 6.
Suppose a or b ≤ . Then (1) satisfying conditions (3) has at most three solutions.Proof. Suppose (3) holds. By symmetry, we may assume a > b . Further assume a is not a perfect power.Taking (2) with ( i, j ) = (2 ,
3) we have ra x ( a x − x + ( − γ ) = sb y ( b y − y + ( − δ ). Lemma 2 shows b y − y ≤ Z so Lemma 1 shows y − y < log(8 · ) / log( b ). For each choice of b ≤ δ ∈ { , } , and y − y < log(8 · ) / log( b ), we see that a x is a factor of b y − y + ( − δ < · ; this is small enough tofactor easily, so we can list all factors a x , hence we know all possible a and x . Lemma 2 bounds a x − x ≤ Z .For each choice of x − x < log(8 · ) / log( a ) and of γ ∈ { , } , we calculate the power of b dividing a x − x +( − γ which gives us the maximal possible value for y , call it y ,max . If y ,max >
0, for each y with1 ≤ y ≤ y ,max and h = gcd( a x − x + ( − γ , b y − y + ( − δ ), we can solve for r = ( b y − y + ( − δ ) / ( a x h )and s = ( a x − x + ( − γ ) / ( b y h ). We have y = ( y − y ) + y and x = ( x − x ) + x . We see that c = | ra x − sb y | . We now check if r ( a x + ( − α ) = s ( b y + ( − β ) for some α, β ∈ { , } . If so, we havevalues a , b , c , r , and s for which (1) has at least three solutions.For each set of at least three solutions found for b ≤ emma 7. No instance of (12) with d = 1 has a fourth solution.Proof. By Lemma 6, we only need to consider a > d = 1. If (1) has a further solution ( x, y ) with y = 2,then hra x = ± hc ± hsb = ± (2 a + ( − v ) ± ( a + ( − v +1 )( a + ( − v ) a , contradicting Lemma 1, which requires x ≥
2. So if d = 1, any fourth solution ( x , y ) to (12) must satisfy x > y > x , y ) ≥ a ≥ c ≤ r + s < · socertainly x > d = 1 and v = 0, we have b = a + 1, rh = a + 2, sh = a −
1, and ch = 2 a + 1, where h = gcd( a + 2 , a − x , y ) = (3 , a + 1) + ( a − b = ( a + 2) a . (19)Considering the solution ( x , y ) we get(2 a + 1) + ( a − b y ≡ a + 2) a . (20)Combining (19) and (20) we find ( a − b y − − ≡ a + 2) a , (21)which requires a | y − a ≡ a ≡ g || a + 2 = b + 1 and let a/ a ; then instead of (21) we can use( b y − − ≡ g +3 a which again requires a | y −
3. So we can write y = 3 + ja for some integer j >
0. For some integers M and M we have(2 a + 1) + ( a − ja ) a + (3 + ja )(3 + ja − a + (3 + ja )(3 + ja − ja − a + (3 + ja )(3 + ja − ja − ja − a + M a ! = M a from which we derive, for some integer M ,(2 − j ) a + (cid:18) − j (cid:19) a = M a (22)so that j = wa/ w ≥
0. If w = 0 then (22) becomes impossible. If w > y ≥ (cid:0) a + 2 (cid:1) a > · when a > d = 1, v = 1, we have b = a −
1, which is equivalent to the previous case after reversing the rolesof a and b . Thus, in every case, no instance of (12) has a fourth solution.7 emma 8. (1) satisfying (3) has at most three solutions.Proof. Using Lemma 6, assume a > b > x 2) and ( i, j ) = (2 , r ( a x + ( − α + γ ) = s ( b y + ( − β + γ ) (23)and ra x ( a x − x − ( − γ + δ ) = sb y ( b y − y − ( − γ + δ ) . (24)Suppose x − x = y − y = 1. Since ( ra, sb ) = 1, from (24) we have a x | b − ( − γ + δ ; therefore, since a > b , we have x = 1, ( − γ + δ = − 1, and a = b + 1. Then b y | a − ( − γ + δ = a + 1 = b + 2 which isimpossible for b > 2. Thus we have max( x − x , y − y ) ≥ 2. By Lemma 2, max( x − x , y − y ) ≤ a > b > x − x , y − y ) = 2 . (25)Taking the ratio of (23) and (24), a x ( a x − x − ( − γ + δ ) a x + ( − α + γ = b y ( b y − y − ( − γ + δ ) b y + ( − β + γ . (26)We rewrite this as( a x − x − ( − γ + δ ) (cid:18) − ( − α + γ a x + ( − α + γ (cid:19) = ( b y − y − ( − γ + δ ) (cid:18) − ( − β + γ b y + ( − β + γ (cid:19) . From this we obtain a x − x − b y − y = a x − x ( − α + γ a x + ( − α + γ − b y − y ( − β + γ b y + ( − β + γ − ( − α + δ a x + ( − α + γ + ( − β + δ b y + ( − β + γ so | a x − x − b y − y | ≤ a x − x a x + ( − α + γ + b y − y b y + ( − β + γ + 1 a x + ( − α + γ + 1 b y + ( − β + γ . Since a x + ( − α + γ ≥ b and b y + ( − β + γ ≥ b − | a x − x − b y − y | ≤ a x − x b + b y − y b − b + 1 b − a x − x , b y − y ) ≥ b , | a x − x − b y − y | < max( a x − x , b y − y ) (cid:18) b + 1 b − b + 1( b − b (cid:19) . Therefore, | a x − x − b y − y | < b − (cid:0) a x − x , b y − y (cid:1) 8o that, since b > (cid:18) − b − (cid:19) b y − y < a x − x < (cid:18) − b − (cid:19) − b y − y (27)and similarly (cid:18) − b − (cid:19) a x − x < b y − y < (cid:18) − b − (cid:19) − a x − x . (28)We now show x ≤ x − x . From (24) and (28) we have a x | b y − y − ( − γ + δ < (cid:18) − b − (cid:19) − a x − x + 1 < a x − x +1 , so x ≤ x − x . Similarly, one can show that y ≤ y − y .We can further show x < x − x . Suppose z = x = x − x . Then (24) gives a z | b y − y − ( − γ + δ so(28) shows a z | b y − y − ( − γ + δ < (cid:18) − b − (cid:19) − a z + 1 < a z . Thus, a z = b y − y − ( − γ + δ and (26) becomes( b y − y − ( − γ + δ b y + ( − β + γ ) = b y ( b y − y + ( − α + γ − ( − γ + δ )which is impossible modulo b . Thus, 1 ≤ x < x − x ; similar arguments show 1 ≤ y < y − y .Recalling (25) we find x = y = 1 and x = y = 3. If ( − γ + δ = 1 then (26) can be rewritten as a ( a − ( − α + γ ) = b ( b − ( − β + γ ). This implies a = b − ( − β + γ and so b = a − ( − α + γ , β = γ and α = γ ,so c = − ( − α r + ( − β s > α = γ = 1 and β = 0. Then r = ( a − /h , s = ( a + 1) /h , and c = (2 a − /h where h = gcd( a − , a + 1) ≤ 3. We see that the case under consideration in this paragraphsatisfies (12) with d = v = 1. By Lemma 7, this cannot lead to a fourth solution.So ( − γ + δ = − 1, and (26) becomes a ( a + 1)( b + ( − β + γ ) = b ( b + 1)( a + ( − α + γ ) . Since gcd( a + 1 , a ± ≤ b + 1 , b ± ≤ 2, we must have a + ( − α + γ | b + ( − β + γ ) , b + ( − β + γ | a + ( − α + γ ) . (29)Note that (27) gives a < b q − b − < b + 2for b ≥ a = b + 1. But then (29) is impossible. x < x < x < x , y < y < y < y c = ( − α r + sb y = ra x + ( − β s = ( − γ ( ra x − sb y ) = ( − δ ( ra x − sb y ) (30)9or some α , β , γ , and δ ∈ { , } . Applying (2) with ( i, j ) = (2 , , , r ( a x − ( − γ a x ) = s ( b y + ( − β + γ ) , (31) r ( a x − ( − α ) = s ( b y − ( − β ) , (32)and r ( a x − ( − α + γ ) = s ( b y + ( − γ b y ) . (33)Taking the ratio of (31) and (32), we obtain a x − ( − γ a x a x − ( − α = b y + ( − β + γ b y − ( − β . (34)Similarly, the ratio of (33) with (32) gives a x − ( − α + γ a x − ( − α = b y + ( − γ b y b y − ( − β , (35)and considering (31) with (33), one gets a x − ( − γ a x a x − ( − α + γ = b y + ( − β + γ b y + ( − γ b y . (36) Lemma 9. For b ≤ and a < · , b σ b ( a ) < .Proof. Let b = p be an odd prime less than 1000. Let k = ⌈ 22 log(10) / log( b ) ⌉ , so p k ≥ . If p σ p ( a ) ≥ then σ p ( a ) ≥ k . Choose α ∈ { , } to minimize the positive integer n such that p | a n + ( − α . Notethat n | ( p − / 2. By definition, p σ p ( a ) | a n + ( − α . Let a be any solution to x n + ( − α ≡ p .Using Hensel’s lifting lemma, we find a unique solution a to the congruence a n + ( − α ≡ p k with a ≡ a mod p . For each prime p < , for each n | ( p − / 2, and for each solution a mod p , calculationsshow that the associated a mod p k exceeds 8 · . In other words, for every prime p < , if p σ p ( a ) ≥ ,then a > · .Now suppose b = p β p β . By definition of σ , b σ b ( a ) = p g p g for some positive integers g and g . Forany given 1 ≤ k ≤ 22 log(10) / log( p ), let k = (cid:24) 22 log(10) − k log( p )log( p ) (cid:25) . Note that if b σ b ( a ) = p g p g ≥ for some a , then there exists 1 ≤ k ≤ 22 log(10) / log( p ) such that k ≤ g and k ≤ g .Suppose b σ b ( a ) = p g p g with k ≤ g and k ≤ g . By definition of σ , there exists n | ( p − / n | ( p − / 2, and α , α ∈ { , } such that p g | a n + ( − α , p g | a n + ( − α , so a n + ( − α ≡ p k , a n + ( − α ≡ p k . For each choice of n | ( p − / 2, we can list all values of a mod p with a n + ( − α ≡ p .Using Hensel’s lifting lemma, we can obtain a complete list of all possible values of a mod p k satisfying10 n + ( − α ≡ p k . Similarly, for each choice of n | ( p − / 2, we can obtain a complete list ofall possible values of a mod p k satisfying a n + ( − α ≡ p k . Thus, for each choice of p , p , k , n | ( p − / n | ( p − / 2, and α , α ∈ { , } , we can obtain every possible a mod p k p k satisfying p k | a n + ( − α , p k | a n + ( − α . Calculations show that each potential a exceeds 8 · . In other words, if b σ b ( a ) ≥ then a > · .Now suppose b = p β p β p β with p < p < p . By definition of σ , b σ b ( a ) = p g p g p g for some positiveintegers g , g , and g . For any given k ≤ 22 log(10) / log( p ), and any k ≤ (22 log(10) − k log( p )) / log( p ),let k = (cid:24) 22 log(10) − k log( p ) − k log( p )log( p ) (cid:25) . If b σ b ( a ) = p g p g p g ≥ for some a , there exist k and k such that k ≤ g , k ≤ g , and k ≤ g . Foreach choice of p , p , p , k , k , n | ( p − / n | ( p − / n | ( p − / α , α , α ∈ { , } , weproceed as in the previous paragraph to obtain every possible a mod p k p k p k satisfying p k | a n + ( − α , p k | a n + ( − α , p k | a n + ( − α . The calculations verify that each possible a exceeds 8 · .Finally, suppose b < is the product of four primes. The same procedure works. Since each b ≤ has four or fewer distinct prime factors, we conclude that if b σ b ( a ) ≥ then a > · .The following two lemmas apply to the following set of solutions:( a, b, c, r, s ; x , y , x , y , . . . , x N , y N ) , N ≥ , x < x < x < · · · < x N , y < y < y < · · · < y N . (37) Lemma 10. If (37) holds, then either x ≤ x − x or y ≤ y − y .Proof. Assume the set of solutions ( a, b, c, r, s ; x , y , x , y , . . . , x N , y N ) satisfies (37), and further assume x > x − x and y > y − y .From (34) we get a x − x − ( − γ + ( − α a x − x − ( − α + γ a x − ( − α = b y − y + ( − β b y − y + ( − β + γ b y − ( − β . (38)If min( a, b ) > 2, we see that | ( − α a x − x − ( − α + γ a x − ( − α | ≤ − = . Similarly, | ( − β b y − y +( − β + γ b y − ( − β | ≤ . In bothcases the value 1 / a (respectively, b ), equals 3, x (respectively, y ) equals 2, and x − x (respectively, y − y ) equals 1. By Lemma 3 of [14], ( a, b ) = 1, so we must have | a x − x − b y − y − ( − γ | < , (39)so the left side of (39) must be zero.But now from (36) we have( − γ a x − x − ( − γ b y − y − − β + γ a x − ( − α + γ b y − ( − α + β a x b y . (40)11ut, from (39), the left side of (40) must be zero, which is impossible since the numerator on the right sideof (40) cannot be zero by Mihailescu’s theorem [9] since x and y both are greater than 2.So we can assume min( a, b ) = 2. We see that | ( − α a x − x − ( − α + γ a x − ( − α | ≤ − = 1. Similarly, we have | ( − β b y − y +( − β + γ b y − ( − β | ≤ 1. In both cases the value 1 is possible only when a (respectively, b ), equals 2, x (respectively, y ) equals 2, and x − x (respectively, y − y ) equals 1. ( a, b ) = 1, so we must have | a x − x − b y − y − ( − γ | < , (41)so the left side of (41) must be zero or one. If the left side of (41) is zero, then again we can use (36) toobtain a contradiction as above. If the left side of (41) equals one, then recall ( a, b ) = 1 and note that a x − x − b y − y = ± a, b ) = 2. Lemma 11. If min( a, b ) > in (37), then x = x − x implies y ≤ y − y , and also y = y − y implies x ≤ x − x .Proof. By symmetry, it suffices to prove that x = x − x implies y ≤ y − y . Assume y > y − y and x = x − x . Since b y − y +( − γ b y − ( − β < 1, we cannot have α = γ in (38), so that a x − x − ( − γ a x − ( − α = 1 ± A ∓ where A = a x = a x − x . Since A ∓ + b y − y +( − γ b y − ( − β < 1, we must have b y − y +( − γ b y − ( − β = A ∓ , so that b y − y + ( − γ | b y − ( − β ). Then, using the elementary divisibility properties of b y ± y ), we see that, since b > 3, we must have y − y | y . Let B = b y − y . Let j = y − y y − y , noting that j is a positive integer. Then 1 + ( − γ B B j (cid:16) − ( − β B j +1 (cid:17) = 2 A (1 ∓ A ) , so that, letting k = min( a, b ) > B ≤ A (1 + k ) − k ) = A (1 + k )2(1 − k ) ≤ A. (42)But from (38) we get B ≥ A − A (1 − A ) ≥ A (1 − k ) ≥ A, contradicting (42). Lemma 12. Suppose a, b ≤ . Then (1) satisfying conditions (4) has at most three solutions.Proof. Assume (4) holds. By symmetry, we may assume a > b . Let b ≤ and assume a is not a perfectpower. Considering (2) with ( i, j ) = (3 , 4) and applying Lemma 3 and Lemma 9, we have b y | b σ b ( a ) ( x − x ) < Z < · , (43)so y < log(8 · ) / log( b ).Choose b ≤ ν ∈ { , } . Considering (31) and noting that ( ra, sb ) = 1, a x must be a divisorof b y + ( − ν . It is easy to factor b y + ( − ν < · , so for each b , y , and ν we obtain every possible a x hence a complete list of possible values for a and its associated x exponent. Lemma 2 gives a bound12 − x < log(8 · ) / log( a ). For each x − x within this bound and each µ ∈ { , } , we can solve r = ( b y + ( − ν ) / ( a x h ) and s = ( a x − x + ( − µ ) /h where h = gcd( a x − x + ( − µ , b y + ( − ν ).Considering (2) with ( i, j ) = (1 , r ( a x + ( − η ) = sb y ( b y − y + ( − θ ) for some η , θ ∈ { , } . We now determine if there is a value η ∈ { , } for which b | a x + ( − η , in which case wedetermine y such that b y || a x + ( − η . Now we see if there is a value θ ∈ { , } for which r ( a x + ( − η ) = sb y ( b y − y + ( − θ ). If so, we have three solutions to (1) with c = | ra x − sb y | .Now apply bootstrapping as outlined above; our calculations show that in each case x or y exceeds8 · , hence there is no fourth solution.We can reformulate (11) as( b kd + ( − u + v b d + ( − u , b, ab d − ( − u + v h , b d + ( − u h , a + ( − v h ; 0 , d, , , , kd ) , (44)where h = gcd( a + ( − v , b d + ( − u ), d and k are positive integers, and if u = 0 then k − v must be odd,or if u = 1 then v = 0. Lemma 13. No member of the infinite class (44) with b > has a fourth solution.Proof. Suppose (1) has three solutions satisfying (44). If r + sb d = ra + s = c , then u = v = 1, contradictingthe conditions of (44). So we must have either sb d − r = c (45)or ra − s = c. (46)Assume (44) has a fourth solution ( x , y ). By Lemma 1, we can assume (4) holds, so that x > , y > kd. (47)By Lemma 12, we may assume b > a ≥ b ( k − d − b ( k − d > · when b > k − d ≥ k − d < 5. For all d and k satisfying this bound, for u, v ∈ { , } , andfor 1000 < b ≤ < b ≤ b > k − d < 4. In particular, theonly pairs ( d, k ) that we need to consider are (3 , , , , , k = 2 implies b d − ( − u = a so that, when k = 2 we must have h ≤ d > k = 2, we certainly have d > r, s ) > . (48)We will now use Lemma 5 to show d | y . When (45) holds, we apply Lemma 5 to the solutions (0 , d ),(2 , kd ), and ( x , y ), noting that (47) gives Conditions (1.) and (3.) of Lemma 5, (45) gives Condition (2.),and, if we assume d > ra, sb ) = 1). Now we can useLemma 5 to get ( k − d | y − d . Similarly, when (46) holds, we can apply Lemma 5 to the solutions(1 , , kd ), and ( x , y ) to get kd | y . In either case, we have d | y . So, in considering the set ofsolutions ( a, b, c, r, s ; 0 , d, , , , kd, x , y ), we can assume d = 1 without loss of generality, noting that wehave reformulated the meanings of b and y . 13hen d = 1 and k = 2, 3, or 4, we can apply the LLL basis reduction method to show that there areno solutions for b ≤ · . There are a handful of b values for which the LLL method fails, and we handlethese by bootstrapping.Using a ≥ b − x , y ) ≥ a we see that y = max( x , y ) implies y ≥ a ≥ b − 1, while x = max( x , y ) implies a a − ( a ( a + 1) + 1) ≤ ra x − c ≤ sb y ≤ ( a + 1) y +1 . In either case, we certainlyhave y > d = 1 and k = 4. Then a ≥ b − b + b − > · for b > · so this case cannot lead to afourth solution.Suppose d = 1 and k = 3, so a = b ∓ b + 1, rh = b ± sh = b ∓ b + 2, and ch = b ∓ b + b ∓ 1, wherewe take the upper sign when u = v = 0 and the lower sign when u = 1 and v = 0. From (44) we have ra x + c ≡ ra + c ≡ sb . (49)When b is even, h is odd and s is even, so ra x + c ≡ ra + c ≡ b . (50)When b ≡ g || a + 1 = sh , we have, noting 2 h , ra x + c ≡ ra + c ≡ g +3 ( b/ . (51)Let n be the least number such that a n ≡ G where G = b when b is odd, G = 2 b when b ≡ G = 2 g +3 ( b/ when b ≡ a x − x ), we find that when b is odd, n = b , and when b is even, n = 2 b . Now from (49), (50),and (51) we see that we have x = 2 + jb where j ≥ b is even, j iseven, although for this case we will not need this.) We have, for some integer M , a x = ( b ∓ b + 1) x = 1 + x ( b ∓ b ) + x ( x − b ∓ b ) + x ( x − x − b ∓ b ) + M ( b ∓ b ) and thus rha x + ch = ( b ± jb )( b ∓ b ) + (2 + jb )(1 + jb )2 ( b ∓ b ) + (2 + jb )(1 + jb ) jb b ∓ b ) + M ( b ∓ b ) ! + ( b ∓ b + b ∓ . Collecting like powers we find (recalling y > rha x + ch = (2 − j ) b + M b = M b where M and M are integers. Thus, j = wb/ w ≥ 0, so that, when w > x ≥ b + b / > · since b > · , contradicting Lemma 1. So w = 0 and x = 2 + 2 b < · ,so that b < · . We apply (8) for each b with 6 · < b ≤ · . The calculations show that for every b in this range, (8) never gives an integral value for y within 25 places of accuracy, hence x = 2 + 2 b cannotlead to a fourth solution. Thus, when d = 1 and k = 3, (44) cannot have a fourth solution.14uppose d = 1, k = 2, u = 1, and v = 0, so a = b + 1, rh = b − sh = b + 2, and ch = b + b + 1. From(44) we have ra x + c ≡ ra + c ≡ sb . Now proceeding as in the case k = 3, we derive x = 2 + jb where j is even when b is even. Write x = 2 + e b + e b with 0 ≤ e < b , noting 2 | e when 2 | b . For someinteger M we have rha x + ch = ( b − e b + e b ) b + (2 + e b + e b )(1 + e b + e b )2 b + M b ) + ( b + b + 1) . Noting (cid:0) x ( x − / (cid:1) − b , and noting y > M and M , rha x + ch = (2 − e ) b + M b = M b so that e ≡ b , so e = 2.Write x = 2 + 2 b + e b with 0 ≤ e . For some integer M we have rha x + ch = ( b − b + e b ) b + (2 + 2 b + e b )(1 + 2 b + e b )2 b + (2 + 2 b + e b )(1 + 2 b + e b )(2 b + e b )6 b + M b ! + ( b + b + 1) . From this we obtain, for some integers M and M , rha x + ch = − e b + M b = M b . Thus e = wb/ w ≥ 0. As before, we show w = 0, so it remains only to deal with x = 2 + 2 b , in which case (8) implies y = 1log( b ) ( x log( b + 1) − log( b + 2) + log( b − b ) ( x log( b ) + x log(1 + 1 /b ) − log(1 + 2 /b ) + log(1 − /b )) . (52)From this one gets y − x = 1log( b ) (cid:18) x log (cid:18) b (cid:19) − log (cid:18) b (cid:19) + log (cid:18) − b (cid:19)(cid:19) . Plugging in x = 2 + 2 b into x log (cid:0) b (cid:1) − log (cid:0) b (cid:1) + log (cid:0) − b (cid:1) and taking Taylor series in terms of z = 1 /b , one can show that the Taylor series 2 − z + 7 z / − z / . . . is alternating, hence,0 < b ) (cid:18) − b (cid:19) < y − x < b ) < y being an integer. So x = 2 + 2 b is not possible in this case, completing the proof thatno fourth solution is possible when d = 1, k = 2, u = 1, v = 0.Suppose d = 1, k = 2, u = 0, and v = 1. Then a = b − b to represent b d ),and reversing the roles of a and b , this becomes identical to the case just completed.Thus (44) does not have a fourth solution. 15 emma 14. (1) satisfying conditions (4) has at most three solutions.Proof. By Lemma 12 we can assume a > b > a x − x + ( − α a x − x − ( − γ a x a x − ( − α = b y − y + ( − β b y − y + ( − β + γ b y − ( − β and so | a x − x − b y − y | ≤ a x − x a x − a x − b y − y b y − b y − . (53)Then the right side of (53) can be bounded as follows: | a x − x − b y − y | ≤ a x − x b + 1 + 1 b + b y − y b − b − , (54)and noting that max( a x − x , b y − y ) ≥ b , | a x − x − b y − y | ≤ max( a x − x , b y − y ) (cid:18) b + 1 b + 1 b + 1 b − b − b (cid:19) so | a x − x − b y − y | < max( a x − x , b y − y ) (cid:18) b − (cid:19) . Since b > (cid:18)(cid:18) − b − (cid:19) b y − y (cid:19) x − x < a < (cid:18) − b − (cid:19) − b y − y ! x − x . (55)Since a > b , this implies x − x ≤ y − y .Since 1000 < b and b y − y < · we must have y − y ≤ 4. We divide this proof into a number ofcases: 2 ≤ x − x < y − y ≤ , ≤ x − x = y − y ≤ x − x = y − y = 2 x − x = 1 , y ≥ y − y x − x = 1 , y < y − y = 3 , x − x = 1 , y < y − y = 2In each case, a key idea is to use (53).If 2 ≤ x − x ≤ y − y ≤ 4, noting that max( a x − x , b y − y ) ≥ b , (54) implies | a x − x − b y − y | ≤ max( a x − x , b y − y ) (cid:18) b + 1 b + 1 b + 1 b − b − b (cid:19) so | a x − x − b y − y | < max( a x − x , b y − y ) (cid:18) b − (cid:19) . b > (cid:18)(cid:18) − b − (cid:19) b y − y (cid:19) x − x < a < (cid:18) − b − (cid:19) − b y − y ! x − x . (56)Suppose 2 ≤ x − x < y − y ≤ 4. Given y − y and x − x with these bounds, consider each b with1000 < b < (8 · ) y − y . Then for each a satisfying (56) with gcd( a, b ) = 1, we consider (2) with ( i, j ) =(3 , 4) and apply Lemma 3 to get b y | b σ b ( a ) ( x − x ) < b σ b ( a ) · . Thus, y < σ b ( a ) + log(8 · ) / log( b ).From (31) a x | b y + ( − β + γ . For each y < σ b ( a ) + log(8 · ) / log( b ) and each β , γ ∈ { , } , let x ,max be the largest power of a dividing b y + ( − β + γ . If x ,max > ≤ x ≤ x ,max and each α ∈ { , } , we set y = y − ( y − y ), r = ( b y − ( − β ) /h , and s = ( a x − ( − α ) /h where h = gcd( a x − ( − α , b y − ( − β ). Let c = ( − α r + sb y . If c = ra x + ( − β s = ( − γ ( ra x − sb y ) thenwe have three solutions to (1). Our calculations show that for ( x − x , y − y ) = (2 , , , a, b, c, r, s ; x , y , x , y , x , y ) =(56744 , , , , , , , , , , (56745 , , , , , , , , , . We apply bootstrapping to these; our calculations show that either there is no fourth solution or else y > · , impossible, so these two sets of three solutions do not extend to a fourth solution.Suppose x − x = y − y = z where z = 3 or 4. From (56) we obtain the impossibility b < a < (cid:18) − b − (cid:19) − z b ≤ (cid:18) b − (cid:19) / b < b + 1 . (57)Suppose x − x = y − y = 2. We consider several subcases.Suppose x ≥ x − x = 2 and y ≥ y − y = 2. Then, since b > 4, (53) implies the impossibility a − b < b + 1 + 1 b + 1 + 1 b − b − < . Suppose x ≥ x − x = 2, 1 = y < y − y = 2. Then, since b > 4, (53) implies the impossibility2 b + 1 ≤ a − b < b + 1 + 1 b + b b − b − < b + 4 . Suppose 1 = x < x − x = 2, y ≥ y − y = 2. Then, since b > 4, (53) implies the impossibility2 a − ≤ a − b < a a − a − b − b − < a + 4 . Suppose 1 = x < x − x = 2, 1 = y < y − y = 2. Then (56) shows that b < a < b q − b − < b (cid:18) b − (cid:19) < b + 2for b > a = b + 1. Now (34) implies( b + 1)(( b + 1) − ( − γ )( b − ( − β ) − ( b + 1 − ( − α )( b + ( − β + γ ) = 0 . (58)17xpanding in powers of b , one gets a polynomial q b + q b + q b + q where each coefficient satisfies | q i | ≤ | q | > 0. No such polynomial can have a zero for b > x − x = 1.Consider first the case x − x = 1 and y ≥ y − y . By Lemmas 10 and 11, we can assume y = y − y ,so that, by Lemma 11 x = x − x = 1, in which case (38) is possible only when α = γ = β , so that wehave (44) with k = 2, u = α , and v = β , which has no fourth solution by Lemma 13.Consider next the case x − x = 1 and y < y − y . Since x ≥ x − x = 1, (53) can be bounded as | a − b y − y | ≤ b + 1 + 1 b + b y − y b − b − . Since y < y − y , we have | a − b y − y | ≤ b y − y (cid:18) b + 1 b + 1 b + 1 b + 1 b − b − b (cid:19) , and therefore | a − b y − y | ≤ b y − y (cid:18) b − (cid:19) . (59)One can now derive (cid:18) − b − (cid:19) b y − y < a < (cid:18) b − (cid:19) b y − y . (60)If y − y = 3 or 4, then for each b with 1000 < b < (8 · ) y − y , each y with y ≤ y − y , and each β, γ ∈ { , } , we can calculate y = y + ( y − y ) and then factor b y + ( − β + γ . Since a x | b y + ( − β + γ ,we now consider each factor a | b y + ( − β + γ that satisfies (60), and calculate the maximal value of x suchthat a x | b y + ( − β + γ ; call this x ,max . For each x up to x ,max , let x = x + 1. For each α ∈ { , } andfor the chosen β , set r = ( b y − ( − β ) /h , s = ( a x − ( − α ) /h , where h = gcd( a x − ( − α , b y − ( − β ).We must have ( − α r + sb y = ra x + ( − β s = ( − γ ( ra x − sb y ). Our calculations now show that theonly possible sets of three solutions belong to the infinite class (44) with x = 1; by Lemma 13, none of theseextends to four solutions.Consider now the case x − x = 1 and 1 = y < y − y = 2. We still have (60) so (cid:16) − b − (cid:17) b < a < (cid:16) b − (cid:17) b . Now a x | b + ( − β + γ but a > (cid:16) − b − (cid:17) b > b + 1, so x = 1.If α = γ then, using (34) and clearing denominators, we get a ( a + ( − α )( b − ( − β ) = ( b + ( − β + γ )( a − ( − α ) . Since gcd( a − ( − α , a + ( − α ) ≤ 2, we must have a − ( − α | b − ( − β ) so a ≤ b + 3. This contradicts (cid:16) − b − (cid:17) b < a .If α = γ = 0 then, using (34) and dividing out a − a ( b − ( − β ) = b + ( − β . (61)Since gcd( b − ( − β , b + ( − β ) ≤ b − ( − β > 2, this is impossible.Therefore α = γ = 1. In this case, using (34) as before and dividing out a + 1, we get (61), so a = b + ( − β b + 1 . So we have a member of the infinite class (44) with d = 1, k = 3, u = β + 1, v = 0; by Lemma 13 this hasno fourth solution. 18 Case (5) x < x < x < x , y = y < y < y c = s − ( − α r = ra x − s = ( − β ( ra x − sb y ) = ( − γ ( ra x − sb y )for some α , β , γ ∈ { , } . Considering (2) with ( i, j ) = (1 , r ( a x + ( − α ) = 2 s , so r = 1 or 2. Applying (2) with ( i, j ) = (1 , 3) and (2 , r ( a x + ( − α + β ) = s ( b y + ( − β ) and ra x ( a x − x + ( − β +1 ) = s ( b y + ( − β +1 ). We find that, for any instance of case (5), we have r ∈ { , } , | a − r, s = r ( a x + ( − α )2 , c = s + ( − α +1 r, b y = 2( a x + ( − α + β ) a x + ( − α − ( − β . (62)Since b y is an integer, we can use the elementary divisibility properties of a x ± x | x when a > 3. And, since (62) holds when x is replaced by x i where 3 < i ≤ N , we have, for a > x | x i , ≤ i ≤ N. (63)Note that (62) corresponds to the infinite family (16). Lemma 15. Suppose a ≤ . Then (1) with conditions (5) has at most three solutions.Proof. Suppose (5) holds and a ≤ x and x − x . Lemma 2 gives a x − x ≤ Z so x − x < log(8 · ) / log( a ). ByLemma 2, s ≤ Z + 1 so a x ≤ Z + 1) + 1 < · + 3, hence x < log(16 · + 3) / log( a ). If a ≥ x | x − x < log(8 · ) / log( a ).Given a , x , x − x , and choosing α , β ∈ { , } , and r ∈ { , } with 2 | a − r , we have (62). Supposefor some choice of a , α , β , x , and x − x , we find that a x +( − α + β ) a x +( − α − ( − β is an integer; it is b y , sowe know b and its associated y (without loss of generality we may assume b is not a perfect power). Wenow have three solutions to (1) with this a , b , c , r , s and choice of signs. We use LLL (then bootstrappingif LLL fails) to show no fourth solutions exist. A few cases with a = 2 and b = (2 x + 1) / Lemma 16. No instance of (5) has four or more solutions.Proof. Suppose (5) holds with N > 3. By Lemma 15 we may assume a > r = 1 if a is odd, r = 2 if a is even. If s ≤ 2, then, considering the solution ( x , y ), we musthave a ≤ 5, contradicting Lemma 15. So we can assume s > 2. We can also assume ra x > 2. And wecertainly can assume that the solutions ( x , y ), . . . , ( x N , y N ) include all solutions to (1) for ( a, b, c, r, s ).Now we can apply Lemma 5 to the solutions ( x , y ), ( x , y ), ( x , y ) to get y | y . (64)Since we have a > a x − x > · when x − x ≥ 3, hence, by Lemma 2, we have x − x ≤ 2. By (63) we see that, in considering the set of solutions ( a, b, c, r, s ; 0 , , x , , x , y , x , y ), wecan assume x = 1 without loss of generality, noting that we are reformulating a , x and x . Now one cansee that the only possibilities satisfying (62) have x = 1 with x − x α β c ≤ r + s < · +2 by Lemma 1, and ± c = ra x − sb y . By Lemma 2, 134000 < a < max( x , y ),so certainly 100 < x . Considering ± c = ra x − sb y modulo a , we have sb y ± c ≡ a . We willexpand y in powers of a ; we have y < · < < a , so we do not need powers of a higher thantwo. By (64), we can assume y = 1 without loss of generality, noting that in doing so we have reformulated b and y . Thus we are considering the following set of solutions: ( a, b, c, r, s ; 0 , , , , x , , x , y ), noting thereformulations of variables as indicated above. Case 1: Suppose x = 1 and x − x = 2. Then β = 0. Using (62) and taking the upper sign when α = 0and the lower sign when α = 1, we have b = 2 a ∓ a + 1 and, for either choice of the parity of a ,( a ∓ 1) + ( a ± b y ≡ ( a ∓ 1) + ( a ± b ≡ a so that ( a ± b y − − ≡ a , so that, for either choice of the parity of a , b y − ≡ a whichrequires a | y − 1, by Lemma 1 of [13]. So, letting y = 1 + ja for some positive integer j , and letting M i be an integer for 1 ≤ i ≤ a ∓ 12 + a ± 12 (2 a ∓ a + 1) y = a ∓ 12 + a ± y (2 a ∓ a ) + y ( y − a ∓ a ) + y ( y − y − a ∓ a ) + y ( y − y − y − a ∓ a ) + 2 M a ! = M a (65)since x > − j ) a ± ja + M a = M a . (66)From this we see that j = 1 + wa/ w ≥ 0. If w = 0 then (66) becomes impossible. And if w > 0, then y > a a > · when a > Case 2: Suppose x = x − x = 1, so that α = β and b = 2 a − ( − α . Considering the solution( x , y ) = (2 , a ,( a − ( − α ) − ( − α ( a + ( − α ) b ≡ a . (67)Let t = 1 if y is odd and α = 0, otherwise let t = 0. Then considering the solution ( x , y ) we have( a − ( − α ) + ( − t ( a + ( − α ) b y ≡ a (68)since x > a + ( − α )( b y − + ( − t + α ) ≡ a , which requires a | y − a , by Lemma 1 of [13]. So we can write y = 1 + e a + e a where0 ≤ e < a .Assume first α = 0. Let M and M be integers. Then using (68) we have a − 12 + a + 12 (cid:18) − y (2 a ) + y ( y − a ) − y ( y − y − a ) + 2 M a (cid:19) = M a . (69)This simplifies to ( − − e ) a = M a M . Thus e = a − 1. So y = 1 + ( a − a + e a . Now from (69) we obtain( − − e ) a = M a for some integer M , so that e = ( wa/ − w ≥ 1. But then y ≥ a − a + (cid:0) a − (cid:1) a > · when a > α = 1. We define M i , 1 ≤ i ≤ a + 12 + a − (cid:18) y (2 a ) + y ( y − a ) + 2 M a (cid:19) = M a . This simplifies to (1 − e ) a = M a . Thus, e = 1. Write y = 1 + a + e a . Then a + 12 + a − y (2 a ) + y ( y − a ) + y ( y − y − a ) + y ( y − y − y − a ) + 2 M a ! = M a . This simplifies to − e a + 23 a = M a (70)so that e = wa/ w ≥ 0. If w = 0 then (70) becomes impossible. And if w > y ≥ a + a a > · since a > y ≤ y − y . If N > 3, applying Lemma 2 we get c ≤ r + sb y ≤ r + sb y − y ≤ ( Z + 1) + ( Z + 1) Z = ( Z + 1) . (71)It follows from the proof of Theorem 2 of [14] (Case 2, equation (76a)), that we must have Z < . c )log(2) + 1 . · · log( Z + 1) log( Z ) log(1 . eZ ) . (72)Using (71) in (72) we get Z < . · , completing a simpler and shorter proof of Theorem 2 of [14]. References [1] M. Bennett, On some exponential equations of S. S. Pillai, Canadian Journal of Mathematics , , no.5, (2001), 897–922. 212] Y. F. Bilu, Y. Bugeaud, M. Mignotte, Catalan’s Equation , book in preparation.[3] Y. Bugeaud, F. Luca, On Pillai’s Diophantine equation, New York J. of Math. , , (2006) 193–217.(electronic)[4] R. K. Guy, C. B. Lacampagne, J. L. Selfridge, Primes at a glance, Math. Comp. , , (1987), 183–202.[5] B. He, A. Togb´e, On the number of solutions of the exponential Diophantine equation ax m − by n = c , Bull. Aust. Math. Soc. , , (2010), 177–185.[6] M. Le, A note on the diophantine equation ax m − by n = k , Indag. Math. (N. S.) , June 1992, 185–191.[7] E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in logarithms of algebraicnumbers. II. , Izv. Math. (2000), 1217–1269.[8] M. Mignotte, A kit on linear forms in three logarithms, in preparation, February 7, 2008.[9] P. Mihailescu, Primary cyclotomic units and a proof of Catalans conjecture, J. Reine Angew. Math. (2004), 167-195.[10] R. Scott, On the Equations p x − b y = c and a x + b y = c z , Journal of Number Theory , , no. 2 (1993),153–165.[11] R. Scott, R. Styer, On p x − q y = c and related three term exponential Diophantine equations withprime bases, Journal of Number Theory , no. 2 (2004), 212–234.[12] R. Scott, R. Styer, On the generalized Pillai equation ± a x ± b y = c , Journal of Number Theory , (2006), 236–265.[13] R. Scott, R. Styer, The generalized Pillai equation ± ra x ± sb y = c , Journal of Number Theory , (2011), 1037–1047.[14] R. Scott, R. Styer The generalized Pillai equation ± ra x ± sb y = c , II. submitted to Journal of NumberTheory .[15] R. Scott, R. Styer The equation | p x ± q y | = c in nonnegative x , y . submitted to Canadian Journal ofMathematics .[16] T. N. Shorey, On the equation ax m − by n = k Nederl. Akad. Wetensch. Indag. Math. (1986), no.3, 353–358.[17] N. Smart, The Algorithmic Resolution of Diophantine Equations , Cambridge University Press, Cam-bridge, 1998.[18] R. Styer, Small two-variable exponential Diophantine equations, Mathematics of Computation ,60