#A64 INTEGERS 18 (2018) APPLYING MODULAR ARITHMETIC TO DIOPHANTINE EQUATIONS

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1 #A64 INTEGERS 18 (2018) APPLYING MODULAR ARITHMETIC TO DIOPHANTINE EQUATIONS Ramy F. Taki ElDin Physics and Engineering Mathematics Deartment, Faculty of Engineering, Ain Shams University, Cairo, Egyt Received: 10/24/17, Acceted: 7/16/18, Published: 7/27/18 Abstract In this aer, exerimental results for resenting the success rates in finding integral solutions to the Diohantine equations of two variables using modular arithmetic are shown. Firstly, and for this urose, the emirical algorithm for obtaining exerimental results is resented for Diohantine equations of olynomial-tye. Using modular arithmetic, the algorithm transforms the Diohantine equation to a linear recurrence relation. This defines a sequence of integers { s } s=1 and the integral solution is achieved once this sequence terminates. The algorithm could be successful in solving olynomial Diohantine equations rovided that a redefined numeral system attern to the base (rime number) for one of the two variables exists. Secondly, without any redefined numeral system attern, the algorithm is adoted for equations with exonential terms. Finally, and for exerimental results, four strategies are roosed for the termination of the sequence { s } s. Success rates are comared by testing the strategies over many olynomial-tye Diohantine equations. The exerimental results are resented in detail and some strategies contribute to high success rates in achieving the integral solutions. 1. Introduction Solving a Diohantine equation refers to finding an integral solution to a olynomial equation in more than one variable. Such equations have attracted researchers throughout a long history. Diohantine equations can be considered over the ring of integers Z, the field of rational numbers Q, or even over any finite field F q. Some Diohantine equations ossess infinite integral solutions, e.g., Pell s equation x 2 dy 2 = 1, where d is a ositive square-free integer. Others ossess only a finite number of integral solutions, e.g., the Mordell curve y 2 = x 3 +k contains only a finite number of integral oints (X, Y ). Many researchers studied Diohantine equations which have no integral solutions, e.g., the work of Lu et al. [5]. The most oular of such equations are Fermat s equation and the Mordell curve y 2 = x 3 + k which have no integral solutions for k = (4n 1) 3 4m 2, where there is no rime 3 (mod 4) such that m, as roved in [1].

2 INTEGERS: 18 (2018) 2 In this aer, Diohantine equations over Z that have ositive integral solutions are considered. The exerimental results of the success rates for solving these equations, using only modular arithmetic, are resented. In his tenth roblem, Hilbert asked about the existence of an algorithm suitable for solving any Diohantine equations. In 1970, Matijasevic [7] roved the non-existence of such an algorithm. Meanwhile, researchers looked for a general algorithm for solving Diohantine equations of secial forms [4]. However, in this research, we aim to test the success rate of a secial algorithm in solving Diohantine equations of general form. In articular, the caability of a modular arithmetic algorithm, in solving a general form of Diohantine equations, is measured by evaluating the success rate of the algorithm in finding integral solutions. As a rearation for the exerimental results, an algorithm using only simle modular arithmetic is resented. The algorithm transforms the Diohantine equation to a linear recurrence relation that generates some integer sequence { s }. Once this sequence terminates, an integral solution is obtained and a success is recorded. The remainder of this aer is organized as follows. In Section 2, we rove the main theorem governing the modular arithmetic algorithm roosed for exerimental results. Ellitic curves are of great imortance and have their alications in crytograhy and integer factorization. Various alications may be found in [2, ch. 7]. Therefore, a greater attention has been given to the Diohantine equation b 2 y 2 + b 1 y + b 0 = a 3 x 3 + a 2 x 2 + a 1 x. In Section 3, examles are used to illustrate the roof that the algorithm can succeed in obtaining an integral solution with y-coordinate, that have a known numeral system attern to some base. Recently, an increasing focus on mixed olynomial-exonential Diohantine equations has been noticed [3, 6, 5]. In Section 3, and without any redefined numeral system attern, the results are adated with an examle to find the integral solutions (X, Z) to Diohantine equations containing exonential terms: A 2z + B z + C = a 3 x 3 + a 2 x 2 + a 1 x. Finally, Section 4 contains our exerimental results. In this section, we aim to score the success ercentage of our modular arithmetic algorithm in solving Diohantine equations of the form y 2 +b 1 y+b 0 = x 3. No redefined numeral system atterns are assumed. Instead, some strategies are being suggested. Particularly, four strategies that may lead to the termination of the sequence { s } are roosed, tested, and their success rates are comared. 2. A Modular Arithmetic Algorithm In this section, we consider the Diohantine equations of the form g(y) = f(x), where g(y) = P m j=0 b jy j 2 Z[y] and f(x) = P n i=1 a ix i 2 Z[x]. One may assume n m without any loss of generality. We resent an emirical modular arithmetic

3 INTEGERS: 18 (2018) 3 algorithm for obtaining ositive integral solutions (X, Y ) to the roosed Diohantine equations. Later on, the success rates of this algorithm are calculated. Definition 1. For an integral solution (X, Y ) to g(y) = f(x) and a fixed rime number, define ` = log (1 + max{x, Y }). For s = 1, 2,..., `, define the unique integers s X (mod s ), s Y (mod s ), s X s (mod ) and s s Y s s (mod ) satisfying 0 ale s, s < s and 0 ale s, s <. Lemma 1. Let h(x) 2 Z[x] be a non-trivial olynomial with a leading coe not divisible by. If h (x) is the reduction of h(x) over F, then cient General case: If gcd(h (x), x 2 Z. x) = 1, then h( ) 6 0 (mod ) for every Secial case: If h(x) = 3a 3 x 2 + 2a 2 x + a 1, then h( ) 6 0 (mod ) for every 2 Z if and only if a 2 2 3a1a3 = 1, where is the Legendre symbol. Proof. Let 2 Z and be its image in F. Assume h( ) 0 (mod ), then (x ) h (x). On the other hand, from Fermat s little theorem, = 0, so (x ) (x x). Thus h( ) 6 0 ( mod ) since otherwise the corimality of h (x) and x x is contradicted. In fact, if gcd(h (x), x x) = 1, then h (x) is irreducible over F or a roduct of irreducible factors each of degree at least 2. Consider the secial case of h(x) = 3a 3 x 2 +2a 2 x+a 1 with - 3a 3. If 3a1a3 6= 1, then there exists 2 Z such that 2 a 2 2 3a 1 a 3 (mod ). By direct substitution, one sees that h( ) 0(mod ), for any 2 Z such that ( a 2 )(3a 3 ) 1 (mod ). This contradicts h( ) 6 0 (mod ) for every 2 Z. Conversely, if there exists 2 Z that satisfies h( ) 0 (mod ) and a2 2 3a1a3 = 1, then, using (3a 3 + a 2 ) 2 a 2 2 3a 1 a 3 (mod ), a contradiction follows since a 2 1 = 2 3a 1 a 3 (3a3 + a 2 ) 2 2 3a3 + a 2 = = 6= 1. a 2 2 Theorem 1. Let (X, Y ) be a ositive integral solution to g(y) = f(x). If is a rime such that gcd(f(x), 0 x x) = 1, then s ( s g 0 ( 1 ) + s )(f 0 ( 1 )) 1 (mod ) ; s = 1, 2,..., `, (1) where the integers s are determined such that 1 = 1 (g( 1) f( 1 )) and s+1 = 1 ( s + N s ( s, s, s, s)), (2)

4 INTEGERS: 18 (2018) 4 for some integer valued function N s ( s, s, s, s). Moreover, (X, Y ) is obtained once the sequence { s } terminates. Proof. From Definition 1, s X s (mod ) imlies that X s s + s s (mod s+1 ). Also, X s+1 (mod s+1 ). Thus s+1 s + s s (mod s+1 ). Since 0 ale s+1 < s+1 and 0 ale s + s s < s + s s = s ( s + 1) ale s = s+1, then s+1 = s + s s. Similarly, s+1 = s + s s. Therefore `X `X `X X = 1 + = s + s s + s+1 s 1 = s+1 + s+1 s 1 Y = 1 + =1 `X =1 =s+1 = s + s s + s+1 `X =s+1 =s+1 s 1 = s+1 + s+1 `X =s+1 is an integral solution to g(y) = f(x). For s = 1, 2,..., `, let g( s ) f( s ) = s s for some integer s. Then s+1 s+1 = g( s + s s ) f( s + s s ) mx X n = b 0 + b j ( s + s s ) j a i ( s + s s ) i j=1 mx jx = b 0 + b js j + j=1 = g( s ) + = s s + = s s + = s s + = s s + =1 mx X j b j j=1 mx jx j=1 =1 mx mx =1 j= mx =1 mx =1 =1 s s! s s! j j j b j j b j mx j= i=1 j s j s j s j s g () ( s ) s s n X i=1 s s f( s ) s s s s b j j P j s nx µ=1 = s s + s s g 0 ( s ) s s f 0 ( s ) + nx i=1 µ=1 nx µ=1 i=µ nx µ=1 a i i s + i=1 ix i a i µ nx i a i µ µ s µs µ! s µ µs f (µ) ( s ) µ! mx =2 s s = s s + s s g 0 ( s ) s s f 0 ( s ) + 2s m X! =2 ix µ=1 nx ix a i nx i=µ g () ( s ) ( 2)s s! µ=1 i µ s µ s µs i µ s µ s µs a i i P µ i µ s i s i µ s µ µs µ i s i µ s µ µs µ s 1 nx s µ µs f (µ) ( s ) µ! µ=2! nx g () µ (µ 2)s s ( s ) f (µ) ( s ). µ! µ=2

5 INTEGERS: 18 (2018) 5 Dividing both sides by s, we get s+1 = s + s g 0 ( s ) s f 0 ( s ) + s m X = s + N s ( s, s, s, s). =2 ( 2)s s! g () ( s ) nx µ=2! s µ (µ 2)s f (µ) ( s ) µ! Reducing Eq. (3) modulo and using the fact that s 1 (mod ) and s 1 (mod ), we get s f 0 ( 1 ) s g 0 ( 1 ) + s (mod ). (4) Using Lemma 1 with h(x) = f 0 (x), we have f 0 ( 1 ) 6 0 (mod ). If X ale Y, it is advantageous to write Eq. (4) as s ( s g 0 ( 1 ) + s )(f 0 ( 1 )) 1 (mod ). Finally, since ` = log (1 + max{x, Y }) log (1 + max{x, Y }) log (1 + X) > log (X), then X < `. But ` X (mod `) and 0 ale ` < ` imlies that ` = X. Similarly, ` = Y. Consequently, ` ` = g( `) f( `) = g(y ) f(x) = 0. Thus, an integral solution (X, Y ) = ( `, `) is achieved if and only if ` = 0, i.e., once the sequence { s } terminates. Corollary 1. The Diohantine equation a b 2 y 2 + b 1 y + b 0 = a 3 x 3 + a 2 x a 1 x ; 2 3a 1 a 3 = 1 (5) (3) has an integral solution (X, Y ) = ( `, `) if and only if order, ` = 0, where we have, in b b b 0 a a a 1 1 (mod ) ; 0 ale 1, 1 < (6) 1 = b b b 0 (a a a 1 1 ) s s + s (2b b 1 ) (3a a a 1 ) 1 (mod ), 0 ale s < (8) s+1 = s + s (2b 2 s +b 1 )+ 2 s s b 2 s (3a 3 2 s +2a 2 s +a 1 )+ 2 s s (3a 3 s +a 2 )+ 3 s 2s a 3 (9) s+1 = s + s s (10) (7) s+1 = s + s s. (11) Proof. Let (X, Y ) be an integral solution to g(y) = b 2 y 2 + b 1 y + b 0 = a 3 x 3 + a 2 x 2 + a 1 x = f(x). From Eq. (2), 1 = g( 1) f( 1 ) = b b b 0 (a a a 1 1 ) 2 Z.

6 INTEGERS: 18 (2018) 6 2 Thus b b b 0 (a a a 1 1 ) and Eq. (6) follows. Formal di erentiation yields f 0 (x) = 3a 3 x 2 + 2a 2 x + a 1, f 00 (x) = 6a 3 x + 2a 2, f 000 (x) = 6a 3, g 0 (y) = 2b 2 y + b 1 and g 00 (y) = 2b 2. From Lemma 1, 3a1a3 = 1 imlies that 3a a a (mod ). Hence, Eq. (8) results from Eq. (1). In this case, Eq. (3) becomes a 2 2 s+1 = s + s g 0 ( s ) + s 2 s 2 g00 ( s ) s f 0 ( s ) + s 2 s 2 f 00 ( s ) + 2s 3 s 6 f 000 ( s ) = s + s (2b 2 s + b 1 ) + 2 s s b 2 ( s (3a 3 2 s + 2a 2 s + a 1 ) + 2 s s (3a 3 s + a 2 ) + 3 s 2s a 3 ). As already shown above, Corollary 1 defines an aroriate algorithm to determine an integral solution to Eq. (5), equivalently ` = 0, once the integers s, for 1 ale s ale ` 1, are redefined. In fact, if we roose an integral solution with y-coordinate ossessing a known numeral system attern to the base, then the numbers s can be easily determined from this attern. This is shown in Lemma 2 below. 3. Diohantine Equations with Exonential Terms In this section, the algorithm defined by Corollary 1 is adated to Diohantine equations with exonential terms. Firstly, such equations are transformed to the form (5) with a redefined numeral system attern to the base for Y. Secondly, the algorithm defined by Corollary 1 is roerly alied. Lemma 2. An integral solution (X, Y ) to Eq. (5) with X ale Y = a(t 1) ( 1), for an integer a such that 0 ale a < and an unknown ositive integer t, is achieved once t = 0. Moreover, 1 = s = a for 1 ale s < t. Proof. Assume an integral solution with X ale Y = a(t 1) ( 1), then ` = log (1 + max{x, Y }) = log (1 + Y ) = log (1 + a(t 1) ( 1) ) ale log (1 + ( 1)(t 1) ) = log ( 1) ( t ) = t. Moreover, since ` Y (mod `) a(t 1) ( 1) (mod `) P t 1 =0 a (mod `) P` 1 =0 a (mod `) and 0 ale `, P` 1 =0 a < `, then ` = P` 1 =0 a.

7 INTEGERS: 18 (2018) 7 Assume ` ale t 1. From Corollary 1 and using 0 = ` = g( `) f( `), we get the following contradiction X` 1 Xt 2 Xt 2 Xt 1 Y = ` = a ale a < a t 1 + a = a = a(t 1) ( 1) =0 =0 =0 =0 = Y. Thus ` = t and the integral solution is achieved once ` = t = 0. For 1 ale s < t, from Definition 1, since s Y a(t 1) ( 1) a(s 1) ( 1) (mod s ) and 1 ale s < s, then s = a(s 1) ( 1). Thus 1 = a. Also, since s a( t 1) ( 1) a( t 1) ( 1) s s a( s 1) ( 1) s a(t 1 + t s ) s a( t s ) a (mod ) and 1 ale a, s <, then s = a. Examle 1. Consider an integral solution to the equation y y = x x x with a y-coordinate of the form 7t 1 2. From Lemma 1, f 0 ( 1 ) 6 0 (mod ) since (456000) 2 + 3(654000) (6) 2 + 3(4) 6 = = = However, it is enough to check that f 0 ( 1 ) 6 0 (mod 7), where 0 ale 1 < 7 is chosen to satisfy Eq. (6). Since Y = 7t 1 2 = 3 P t 1 =0 7, then a = 1 = s = 3 and Eq. (6) reduces to (mod 7). It is easily seen that 1 = 4 and f 0 (4) (mod 7). From Eq. (7), 1 = Now, equations (8), (9), (10) and (11) are s (5 s + 6 s ) (mod 7) s+1 = s + s (2 s )+ s 2 7s s (3 s s )+ 2 s 7s (3 s )+ 3 s 72s s+1 = s + s 7 s s+1 = s + s 7 s. Calculations are summarized in Table 1. From Lemma 2, the integral solution is ( 11, 11) = ( , ) since 11 = 0. Theorem 2. An integral solution to a A 2z + B z + C = a 3 x 3 + a 2 x a 1 x ; 2 3a 1 a 3 = 1 (12) is (X, Z) = ( Z, Z). It is obtained once Z = 0 using Eqs. (7), (8), (9), (10) and (11) with b 2 = A, b 1 = 2A + B, b 0 = A + B + C, 1 = s = 1 for 1 ale s < Z and 1 is such that a a a 1 1 C (mod ).

8 INTEGERS: 18 (2018) 8 s s s s s s s s s s s s Table 1: Calculations for finding an integral solution to y y = x x x. Proof. We have (X, Z) is an integral solution to Eq. (12) if and only if A 2Z + B Z + C = a 3 X 3 + a 2 X 2 + a 1 X. The latter is true if and only if A( Z 1) 2 + (2A + B)( Z 1) + A + B + C = a 3 X 3 + a 2 X 2 + a 1 X. Equivalently, (X, Y ) is an integral solution to Eq. (5) with Y = Z 1 = ( 1) P Z 1 =0. From Lemma 2 and Corollary 1, this solution is achieved once ` = Z = 0. Hence, X = ` = Z. Comaring Y with the form in Lemma 2, we deduce that 1 = s = 1 for s = 1, 2,..., Z 1. Finally, Eq. (6) with 1 = 1 reduces to a a a 1 1 C (mod ). Remark The condition a2 2 3a1a3 = 1 in Corollary 1 and Theorem 2 is su cient but not necessary for f 0 ( 1 ) 6 0 (mod ), where 1 is a solution to Eq. (6). 2. The equations can be modified to find an integral solution (X, Z) to An 2z + Bn z + C = a 3 x 3 + a 2 x 2 + a 1 x, where n is any ositive integer. In articular, (X, Z) is an integral solution if and only if (X, Y ) is an integral solution to Eq. (5), with b 2 = A, b 1 = 2A + B, b 0 = A + B + C and ZX 1 Y = n Z 1 = (n 1) n. Equations (7), (9), (10) and (11) are to be modified by relacing by n. Moreover, calculations in Eqs. (6) and (8) are carried over the ring Z/nZ. Once ` = 0, the integral solution is X = ` and Z = `. 3. Using Theorem 1, Theorem 2 may be generalized to Diohantine equations of the form g( z ) = f(x). =0 Examle 2. Consider an integral solution to the equation 11 2z = x 3 + 2x x. Equivalently, from Theorem 2, the algorithm of Corollary 1 is alied to y 2 + 2y = x 3 + 2x x with Y = 11 Z 1. Calculations are summarized in Table 2. Since 10 = 0, an integral solution is X = 10 = and Z = ` = 10.

9 INTEGERS: 18 (2018) 9 s s s s s s s s s s s s Table 2: Calculations for finding an integral solution to 11 2z = x 3 +2x x. 4. Four Strategies and Exerimental Results As mentioned above, no general algorithm is suitable for solving all Diohantine equations. However, in Corollary 1, we showed a modular arithmetic algorithm that succeeded with all Diohantine equations (5) that have a redefined numeral system attern for Y. Also, by Theorem 2, the algorithm succeeded in solving all mixed olynomial-exonential Diohantine equations (12). As we mentioned earlier, our aim is to test a secial algorithm by alying it on general form Diohantine equations. Hence, the olynomial-tye Diohantine equations (5) should be considered without any known numeral system attern. However, to achieve an integral solution, some data seem to be missing. The missing data can be dealt with in a variety of methods that lead to comaring success rates on a samle of Diohantine equations. An integral solution to Eq. (5) is achieved once ` = 0, where the sequence { s } is determined by Eqs. (7) and (9). However, the termination of the sequence deends on the choices of s and s, for 1 ale s < `. Moreover, Eq. (8) is a linear relation between s and s. Therefore, the missing data for achieving an integral solution can be handled by suggesting some strategies. A good strategy is the one that redicts the values of s and s, with resecting Eq. (8), so that the sequence { s } terminates. Since we are roosing a comarison for success rates, some strategies that may lead to an integral solution to Eq. (5) are suggested. In fact, many strategies may be lanned for the goal of ` = 0. However, for comarison uroses, we roose only four strategies that may suit some secial Diohantine equations in the form of (5). Furthermore, the success ercentages of these strategies, in achieving an integral solution, are comuted and comared. Strategies are suggested for Diohantine equations (5) satisfying the following conditions: 1. b 2, b 1, a 3, a 2, a b 1 is odd. 3. X is even if a 1 is odd.

10 INTEGERS: 18 (2018) X is odd if a 3 + a 1 is odd. With these conditions and for = 2, Eq. (8) has the form s + s s (mod 2) ; s, s 2 {0, 1}. (13) Moreover, equations (7) and (9), used in determining the sequence { s }`s=1, have the following forms: 1 = 1 (b 2 + b 1 ) + b 0 1 (a 3 + a 2 + a 1 ) 2 s+1 = s + s b 2 (2 s + 2 s ) + b 1 s a 3 (3 2 s + 3 s 2 s + 2 2s ) + a 2 (2 s + 2 s ) + a 1 2 From Corollary 1, ( `, `) is an integral solution if and only if ` = 0. For this goal, some strategies may be designed deending on the sign and magnitude of s of the current ste and revious stes as well. However, we roose strategies deending only on the sign of the current ste s as follows: Strategy 1: If s > 0, set s = 0 and evaluate s using Eq. (13). If s < 0, set s = 1 and evaluate s using Eq. (13). Strategy 2: If s > 0, set s = 1 and evaluate s using Eq. (13). If s < 0, set s = 0 and evaluate s using Eq. (13). Strategy 3: If s > 0, set s = 0 and evaluate s using Eq. (13). If s < 0, set s = 0 and evaluate s using Eq. (13). Strategy 4: If s > 0, set s = 1 and evaluate s using Eq. (13). If s < 0, set s = 1 and evaluate s using Eq. (13). Examle 3. Consider an integral solution to y 2 + y = x 3 + 2x. Clearly, the x-coordinate should be odd, i.e., X 1 1 (mod 2). If a solution is intended with odd y-coordinate, i.e., Y 1 1 (mod 2), Strategy 2 leads to the integral solution (2175, ). Calculations are summarized in Table 3, where 17 = 0 is observed and the integral solution is deduced immediately..

11 INTEGERS: 18 (2018) 11 s s s s s s s s s s s s Table 3: Calculations for finding an integral solution to y 2 + y = x 3 + 2x using Strategy 2. Over a samle of Diohantine equations, we aim to comare the success ercentage of each strategy in obtaining integral solutions. Recently, Diohantine equations with integral solutions of less than 10 digits are not of any interest, as these kinds of equations can be solved, almost immediately, with a trivial brute force search. However, for comarison uroses, the chosen random samle of Diohantine equations does not need to have large integral solutions. The random samle consists of some Diohantine equations of the form y 2 + b 1 y + b 0 = x 3 that have integral solutions with odd X, b 1 = 1, 3, 5 and b 0 < Precisely, the samle consists of all the Diohantine equations that have been solved successfully through at least one of the four strategies. The coe cients b 1 and b 0 of each equation are listed in Table 4. Moreover, the integral solution and the e ective strategies are recorded next to each equation. In Table 5, we evaluate the ratios of the number of equations solved successfully by each strategy comared to the total number of tested equations, for b 1 = 1, 3 and 5. It was observed that strategies 1, 2, 3 and 4 succeeded to solve 32.5%, 65%, 55% and 39% of the listed equations, resectively. b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg (1,99) 2, (1,98) 2, (19,127) 1, (1,98) 2, (1,97) 2, (1,97) 2, (1,97) 2, (19,127) 1, (1,96) 2, (19,127) 1, (1,96) 2, (19,126) 1, (1,96) 2, (19,126) 1, (1,95) 2, (19,126) 1, (1,95) 2, (1,94) 2, (1,95) 2, (1,94) 2, (1,93) 2, (1,94) 2, (1,93) 2, (1,92) 2, (1,93) 2, (1,92) 2, (1,91) 2, (1,92) 2, (1,91) 2, (1,90) 2, (1,91) 2, (1,90) 2, (1,89) 2, (1,90) 2, (1,89) 2, (1,88) 2, (1,89) 2, (1,88) 2, (1,87) 2, (1,88) 2, (1,87) 2, (1,86) 2, (1,87) 2, (1,86) 2, (1,85) 2, (1,86) 2, (1,85) 2, (21,127) 1, (1,85) 2, (1,84) 2, (1,84) 2, (1,84) 2, (21,127) 1, (1,83) 2,3 Table 4 Continued on next age

12 INTEGERS: 18 (2018) 12 Continued from revious age b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg (21,127) 1, (1,83) 2, (21,126) 1, (1,83) 2, (21,126) 1, (1,82) 2, (1,82) 2, (1,82) 2, (39,255) 1, (21,126) 1, (1,81) 2, (1,81) 2, (1,81) 2, (1,80) 2, (1,80) 2, (1,80) 2, (1,79) 2, (1,79) 2, (1,79) 2, (39,255) 1, (1,78) 2, (1,78) 2, (1,78) 2, (39,254) 1, (1,77) 2, (1,77) 2, (1,77) 2, (39,255) 1, (1,76) 2, (1,76) 2, (1,76) 2, (39,254) 1, (1,75) 2, (1,75) 2, (1,75) 2, (1,74) 2, (1,74) 2, (1,74) 2, (1,73) 2, (39,254) 1, (1,73) 2, (1,72) 2, (1,73) 2, (1,72) 2, (1,71) 2, (1,72) 2, (1,71) 2, (1,70) 2, (1,71) 2, (1,70) 2, (1,69) 2, (1,70) 2, (1,69) 2, (1,68) 2, (1,69) 2, (1,68) 2, (1,67) 2, (1,68) 2, (1,67) 2, (1,66) 2, (1,67) 2, (1,66) 2, (23,127) 1, (1,66) 2, (1,65) 2, (1,65) 2, (1,65) 2, (23,127) 1, (1,64) 2, (1,64) 2, (1,64) 2, (23,126) 1, (23,127) 1, (1,63) 1,2,3, (1,63) 1,2,3, (1,63) 1,2,3, (3,63) 1, (3,63) 1, (3,63) 1, (23,126) 1, (5,63) 1, (5,63) 1, (5,63) 1, (1,62) 1,2,3, (1,62) 1,2,3, (1,62) 1,2,3, (3,62) 1, (3,62) 1, (3,62) 1, (5,62) 1, (23,126) 1, (5,62) 1, (1,61) 2, (5,62) 1, (1,61) 2, (7,63) 1, (1,61) 2, (7,63) 1, (1,60) 2, (7,63) 1, (1,60) 2, (7,62) 1, (1,60) 2, (7,62) 1, (1,59) 2, (7,62) 1, (1,59) 2, (1,58) 2, (1,59) 2, (1,58) 2, (9,63) 1, (1,58) 2, (9,63) 1, (1,57) 2, (1,57) 2, (1,57) 2, (9,62) 1, (9,63) 1, (1,56) 2, (1,56) 2, (1,56) 2, (9,62) 1, (1,55) 2, (9,62) 1, (1,55) 2, (1,54) 2, (1,55) 2, (1,54) 2, (1,53) 2, (1,54) 2, (1,53) 2, (1,52) 2, (1,53) 2, (1,52) 2, (11,63) 1, (1,52) 2, (11,63) 1, (1,51) 2, (11,63) 1, (1,51) 2, (11,62) 1, (1,51) 2, (11,62) 1, (1,50) 2, (11,62) 1, (1,50) 2, (1,49) 2, (1,50) 2, (1,49) 2, (1,48) 2, (1,49) 2, (1,48) 2, (1,47) 2, (1,48) 2, (1,47) 2, (1,46) 2, (1,47) 2, (1,46) 2, (1,45) 2, (1,46) 2, (1,45) 2, (1,44) 2, (1,45) 2, (1,44) 2, (13,63) 1, (1,44) 2, (1,43) 2, (1,43) 2, (1,43) 2, (13,63) 1, (1,42) 2, (13,63) 1, (1,42) 2, (13,62) 1, (1,42) 2, (13,62) 1, (1,41) 2, (1,41) 2, (1,41) 2, (1,40) 2, (13,62) 1, (1,40) 2, (1,39) 2, (1,40) 2, (1,39) 2, (1,38) 2, (1,39) 2, (1,38) 2, (1,37) 2, (1,38) 2, (1,37) 2, (1,36) 2, (1,37) 2, (1,36) 2, (1,35) 2, (1,36) 2, (1,35) 2, (1,34) 2, (1,35) 2, (1,34) 2, (1,33) 2, (1,34) 2, (1,33) 2, (1,32) 2, (1,33) 2, (1,32) 2, (25,127) 1, (1,32) 2, (1,31) 1,2,3, (1,31) 1,2,3, (1,31) 1,2,3, (3,31) 1, (3,31) 1, (3,31) 1, (1,30) 1,2,3, (1,30) 1,2,3, (1,30) 1,2,3, (3,30) 1, (3,30) 1, (3,30) 1, (5,31) 1, (5,31) 1, (1,29) 2, (1,29) 2, (1,29) 2, (5,31) 1, (25,127) 1, (5,30) 1, (1,28) 2, (1,28) 2, (1,28) 2, (5,30) 1, (5,30) 1, (15,63) 1, (1,27) 2, (1,27) 2, (25,126) 1, (33,191) (15,63) 1, (1,27) 2, (1,26) 2, (1,26) 2, (1,26) 2, (15,63) 1, (33,190) (15,62) 1, (7,31) 1, (7,31) 1, (7,31) 1, (1,25) 2, (1,25) 2, (1,25) 2,3 Table 4 Continued on next age

13 INTEGERS: 18 (2018) 13 Continued from revious age b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg (25,127) 1, (15,62) 1, (33,189) (1,24) 2, (7,30) 1, (7,30) 1, (7,30) 1, (1,24) 2, (1,24) 2, (1,23) 2, (25,126) 1, (1,23) 2, (15,62) 1, (1,23) 2, (1,22) 2, (1,22) 2, (1,22) 2, (1,21) 2, (1,21) 2, (1,21) 2, (1,20) 2, (1,20) 2, (1,20) 2, (1,19) 2, (1,19) 2, (1,19) 2, (1,18) 2, (25,126) 1, (1,18) 2, (9,31) 1, (33,190) (33,189) (1,17) 2, (1,18) 2, (1,17) 2, (33,188) (1,17) 2, (9,31) 1, (1,16) 2, (1,16) 2, (1,16) 2, (9,30) 1, (9,31) 1, (1,15) 1,2,3, (1,15) 1,2,3, (1,15) 1,2,3, (9,30) 1, (3,15) 1, (3,15) 1, (3,15) 1, (1,14) 1,2,3, (1,14) 1,2,3, (1,14) 1,2,3, (3,14) 1, (9,30) 1, (3,14) 1, (1,13) 2, (3,14) 1, (1,13) 2, (1,12) 2, (35,207) (35,206) (37,223) 1, (1,13) 2, (1,12) 2, (5,15) 1, (1,12) 2, (1,11) 2, (1,11) 2, (1,11) 2, (5,15) 1, (1,10) 2, (5,15) 1, (1,10) 2, (5,14) 1, (1,10) 2, (5,14) 1, (1,9) 2, (1,9) 2, (1,9) 2, (1,8) 2, (5,14) 1, (1,8) 2, (1,7) 1,2,3, (1,8) 2, (1,7) 1,2,3, (1,6) 1,2,3, (1,7) 1,2,3, (1,6) 1,2,3, (3,7) 1, (1,6) 1,2,3, (3,7) 1, (1,5) 2, (3,7) 1, (1,5) 2, (3,6) 1, (1,5) 2, (3,6) 1, (1,4) 2, (1,4) 2, (1,4) 2, (1,3) 1,2,3, (3,6) 1, (1,3) 1,2,3, (1,2) 1,2,3, (1,3) 1,2,3,4 3-9 (1,2) 1,2,3,4 5-5 (1,1) 1,2,3,4 1-7 (5,11) 1,2,3,4 3-5 (5,10) 1,2,3,4 5-1 (5,9) 1,3 1-5 (1,2) 1,2,3,4 3-3 (1,1) 1,2,3,4 5 1 (1,0) 1,2,3,4 1-1 (1,1) 1,2,3,4 3 1 (1,0) 1,2,3,4 5 3 (3,3) 2,4 1 1 (1,0) 1,2,3,4 3 9 (3,3) 2, (3,2) 2,4 1 7 (3,4) 1,2,3, (5,9) 1, (5,8) 1, (3,3) 2, (3,2) 2, (3,1) 1,2,3, (3,2) 2, (3,1) 1,2,3, (3,0) 1,2,3, (3,1) 1,2,3, (3,0) 1,2,3, (39,241) (3,0) 1,2,3, (39,242) (7,15) 2, (39,243) (5,8) 1, (7,14) 2, (5,9) 1, (7,16) 2, (11,33) (7,17) 2, (15,56) 1, (7,13) 2, (5,8) 1, (7,15) 2, (5,1) 1, (15,57) 1, (7,14) 2, (5,0) 1, (7,16) 2, (5,1) 1, (7,12) 2, (7,15) 2, (5,0) 1, (11,32) (5,1) 1, (7,13) 2, (137,1601) (5,0) 1, (11,33) (7,11) 2, (7,14) 2, (7,12) 2, (23,107) 2, (7,13) 2, (23,108) 2, (7,10) 2, (23,109) 2, (7,11) 2, (7,9) 1,2,3, (15,56) 1, (11,32) (7,8) 1,2,3, (135,1568) (7,10) 2, (7,7) 2, (7,12) 2, (7,9) 1,2,3, (7,6) 2, (11,33) (7,8) 1,2,3, (7,5) 2, (7,11) 2, (7,7) 2, (15,53) 2, (7,10) 2, (7,6) 2, (7,4) 2, (7,9) 1,2,3, (15,54) 2, (7,3) 2, (7,8) 1,2,3, (7,5) 2, (7,2) 2, (11,32) (7,4) 2, (7,1) 1,2,3, (7,7) 2, (7,3) 2, (7,0) 1,2,3, (15,55) 2, (7,2) 2, (9,17) 1, (7,6) 2, (7,1) 1,2,3, (17,65) (7,5) 2, (7,0) 1,2,3, (9,16) 1, (7,4) 2, (9,17) 1, (15,52) 2, (7,3) 2, (15,53) 2, (17,64) (7,2) 2, (9,16) 1, (15,51) 2, (7,1) 1,2,3, (17,65) (63,497) 1, (7,0) 1,2,3, (15,52) 2, (15,50) 2, (15,54) 2, (63,498) (9,1) 1, (9,17) 1, (15,51) 2, (9,0) 1, (9,16) 1, (17,64) (15,49) 1,2,3, (15,53) 2, (9,1) 1, (15,48) 1,2,3, (63,499) (15,50) 2, (15,47) (15,52) 2, (9,0) 1, (79,699) (17,65) (15,49) 1,2,3, (13,33) (15,51) 2, (15,48) 1,2,3, (11,17) 1, (9,1) 1, (79,700) (11,16) 1,3 Table 4 Continued on next age

14 INTEGERS: 18 (2018) 14 Continued from revious age b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg (9,0) 1, (11,17) 1, (13,32) (17,64) (13,33) (15,46) (15,50) 2, (15,47) (15,45) (15,49) 1,2,3, (11,16) 1, (15,44) (79,701) (13,32) (15,43) (15,48) 1,2,3, (15,46) (11,1) 1, (11,17) 1, (15,45) (11,0) 1, (11,16) 1, (15,44) (15,42) (13,33) (11,1) 1, (15,41) (15,47) (11,0) 1, (63,496) 1, (13,32) (15,43) (15,40) (15,46) (15,42) (41,257) (15,45) (63,497) 1, (77,672) (11,1) 1, (15,41) (15,39) (11,0) 1, (77,673) (15,38) (15,44) (15,40) (15,37) (15,43) (15,39) (13,17) 1, (63,498) (15,38) (13,16) 1, (15,42) (13,17) 1, (15,36) (15,41) (13,16) 1, (15,35) (15,40) (15,37) (15,34) (15,39) (15,36) (41,256) (13,17) 1, (15,35) (15,33) 2, (15,38) (41,257) (13,1) 1, (13,16) 1, (15,34) (15,32) 2, (15,37) (15,33) 2, (13,0) 1, (15,36) (13,1) 1, (15,31) 2, (15,35) (13,0) 1, (23,97) 1, (15,34) (15,32) 2, (19,65) (13,1) 1, (15,31) 2, (15,30) 2, (13,0) 1, (79,699) (79,698) (15,33) 2, (15,30) 2, (15,29) 2, (15,32) 2, (19,65) (27,129) (79,700) (15,29) 2, (19,64) (15,31) 2, (23,97) 1, (15,28) 2, (15,30) 2, (15,28) 2, (23,96) 1, (15,29) 2, (63,496) 1, (15,27) 2, (63,497) 1, (15,27) 2, (15,26) 2, (15,28) 2, (19,64) (15,25) 2, (19,65) (41,256) (27,128) (41,257) (15,26) 2, (15,24) 2, (15,27) 2, (27,129) (15,23) 2, (23,97) 1, (23,96) 1, (15,22) 2, (15,26) 2, (15,25) 2, (15,21) 2, (19,64) (15,24) 2, (15,20) 2, (15,25) 2, (15,23) 2, (15,19) 2, (15,24) 2, (15,22) 2, (15,18) 2, (15,23) 2, (15,21) 2, (15,17) 1,2,3, (23,96) 1, (15,20) 2, (47,315) (15,22) 2, (27,128) (15,16) 1,2,3, (15,21) 2, (15,19) 2, (15,15) 2, (27,129) (15,18) 2, (15,14) 2, (15,20) 2, (47,316) (15,13) 2, (15,19) 2, (15,17) 1,2,3, (15,12) 2, (47,317) (15,16) 1,2,3, (15,11) 2, (15,18) 2, (15,15) 2, (15,10) 2, (15,17) 1,2,3, (15,14) 2, (15,9) 2, (15,16) 1,2,3, (539,12512) (15,8) 2, (41,256) (15,13) 2, (15,7) 2, (15,15) 2, (15,12) 2, (15,6) 2, (539,12513) (15,11) 2, (15,5) 2, (15,14) 2, (15,10) 2, (15,4) 2, (27,128) (15,9) 2, (15,3) 2, (15,13) 2, (15,8) 2, (15,2) 2, (15,12) 2, (15,7) 2, (15,1) 1,2,3, (15,11) 2, (15,6) 2, (15,0) 1,2,3, (15,10) 2, (15,5) 2, (31,160) 2, (15,9) 2, (15,4) 2, (63,494) (15,8) 2, (15,3) 2, (17,33) 1, (15,7) 2, (15,2) 2, (31,159) (15,6) 2, (15,1) 1,2,3, (17,32) 1, (15,5) 2, (15,0) 1,2,3, (31,158) (15,4) 2, (31,161) 2, (255,4069) (15,3) 2, (63,495) (31,157) (15,2) 2, (31,160) 2, (63,493) (15,1) 1,2,3, (17,33) 1, (35,193) 1, (15,0) 1,2,3, (17,32) 1, (31,156) (31,162) (31,159) (21,65) (63,496) (255,4070) (21,64) (31,161) 2, (31,158) (17,1) 1, (17,33) 1, (63,494) (17,0) 1, (17,32) 1, (31,157) (31,155) (31,160) 2, (21,65) (35,192) 1, (255,4071) (17,1) 1, (31,154) 2 Table 4 Continued on next age

15 INTEGERS: 18 (2018) 15 Continued from revious age b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg b 1 b 0 (X, Y ) Strg (31,159) (17,0) 1, (63,492) (63,495) (21,64) (19,33) 1, (31,158) (31,156) (31,153) (17,1) 1, (35,193) 1, (19,32) 1, (17,0) 1, (31,155) (287,4859) (21,65) (35,192) 1, (25,97) 1, (31,157) (63,493) (31,152) (21,64) (31,154) (25,96) 1, (31,156) (19,33) 1, (31,151) (35,193) 1, (287,4860) (63,491) (63,494) (19,32) 1, (31,150) (31,155) (31,153) (31,149) (287,4861) (25,97) 1, (19,1) 1, (19,33) 1, (25,96) 1, (19,0) 1, (19,32) 1, (31,152) (29,129) (35,192) 1, (63,492) (31,148) (31,154) (31,151) (29,128) (25,97) 1, (31,150) (31,147) (31,153) (19,1) 1, (63,490) (25,96) 1, (19,0) 1, (153,1888) 1, (63,493) (31,149) (23,65) (31,152) (29,129) (31,146) (31,151) (31,148) (23,64) (19,1) 1, (63,491) (75,641) (19,0) 1, (153,1889) 1, (21,33) 1, (31,150) (29,128) (31,145) (31,149) (31,147) (21,32) 1, (63,492) (23,65) (31,144) (29,129) (23,64) (63,489) (31,148) (31,146) (31,143) (29,128) (21,33) 1, (65,513) (23,65) (21,32) 1, (31,142) (23,64) (31,145) (75,640) (31,147) (63,490) (31,141) (21,33) 1, (31,144) (21,1) 1, (21,32) 1, (31,143) (21,0) 1, (31,146) (75,641) (63,488) (63,491) (31,142) (31,140) (31,145) (21,1) 1, (31,139) (31,144) (21,0) 1, (27,97) 1, (31,143) (63,489) (65,512) (21,1) 1, (31,141) (27,96) 1, (21,0) 1, (31,140) (63,490) (65,513) (31,142) (27,97) 1, (31,141) 2 Table 4: Successful strategies in solving y 2 + b 1 y + b 0 = x 3 with odd X. Strategy 1 Strategy 2 Strategy 3 Strategy 4 b 1 = 1 102/ / / /316 b 1 = 3 103/ / / /315 b 1 = 5 102/ / / /312 Table 5: Strategies success ratios for di erent b 1 values. 5. Conclusion A modular arithmetic algorithm is used to test the success rates of di erent strategies in solving olynomial-tye Diohantine equations. We have roved the success of the algorithm in solving:

16 INTEGERS: 18 (2018) 16 Diohantine equations of the form b 2 y 2 + b 1 y + b 0 = a 3 x 3 + a 2 x 2 + a 1 x with y-coordinate having a known numeral system attern to the base. Diohantine equations of the form A 2z + B z + C = a 3 x 3 + a 2 x 2 + a 2 x. As an integral solution is achieved once the sequence { s } terminates, this suggests that many strategies may lead to a terminated sequence. Four strategies are roosed and their success rates are comared over a samle of Diohantine equations. For simlicity, the roosed strategies rely only on the sign of s for the current ste. Some strategies yield a fairly good success rate. For instance, Strategy 2 succeeded in solving almost 65% of the Diohantine equations listed in Table 4. For future research and better results, other strategies, deending on the signs and magnitudes of the numbers s of the current ste and revious ones, may be suggested and their success rates would be examined and comared. Acknowledgement. The author is very grateful to the anonymous reviewers and the Managing Editor, Professor Bruce Landman, for their detailed comments that have imroved the quality of this aer. References [1] T. M. Aostol, Introduction to Analytic Number Theory, Sringer-Verlag, New York, [2] R. Crandall and C. Pomerance, Prime Numbers: A Comutational Persective, Sringer, New York, [3] L. Hajdu and I. Pink, On the Diohantine equation a + x b = y n, J. Number Theory 143 (2014), [4] C. Hering, On the Diohantine equations ax 2 + bx + c = c 0 c y 1 1 cyr r, Al. Algebra Engrg. Comm. Comut. 7 (1996), [5] Y. Lu and X. Li, A note on the equation x y + y z = z x, J. Inequal. Al. (2014), 2014:170. [6] F. Lucaa and G. Soydan, On the Diohantine equation 2 m + nx 2 = y n, J. Number Theory 132 (2012), [7] Yu. V. Matiyasevich, What can and cannot be done with Diohantine roblems, Proc. Steklov Inst. Math. 275 (2011),