Diophantine Equations. Elementary Methods

Transcription

1 International Mathematical Forum, Vol. 12, 2017, no. 9, HIKARI Ltd, Diophantine Equations. Elementary Methods Rafael Jakimczuk División Matemática, Universidad Nacional de Luán Buenos Aires, Argentina Copyright c 2017 Rafael Jakimczuk. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this note we are interested in some diophantine equations to the form h k x r = k h+1 x r h+1 h+1. Some of these diophantine equations are well-known and our methods of solution are different and very elementary. Mathematics Subect Classification: 11A99, 11B99 Keywords: Diophantine equations, elementary methods 1 Introduction and Main Results In this note we are interested in some diophantine equations to the form k x r = k h+1 x r h+1 h+1 1) where h 2, the coefficients k = 1,..., h + 1) are integers different of zero and the exponents r 2 = 1,..., h + 1) are positive integers. Some of these diophantine equations are well-known see [1] and [2]) and our methods of solution are different and very elementary. Let us consider a solution x 1, x 2,..., x h, x h+1 ) to equation 1) the x = 1,..., h + 1) are integers). If we multiply both sides of equation 1) by E L, where E is an integer different of zero and L is

2 430 Rafael Jakimczuk the least common multiple lcm)of the exponents r = 1,..., h + 1), then we obtain the solution ) x 1 E L r1, x 2 E L r2,..., x h E L rh, x h+1 E L r h+1 3) The solution 3) will be called derivada solution of 1). Note that solution 2) is derivada solution of solution 2) if we put E = 1. Clearly, from 3) we can obtain 2) by common factor. If a set of solutions of equation 1) contain at least one derivada solution of each solution of equation 1) we shall call this set of solutions a complete system of solutions to equation 1). Note that from a complete system of solutions we can obtain all solution to the equation by common factor. This method if not very different to consider the set of primitive solutions to, for example, the equation x 2 + y 2 = z 2 and to obtain the rest of the solutions by multiplication of the primitive solutions. If we consider a certain subset S of solutions to equation 1) then a complete system of solutions in relation to S is a subset of S that contain at least a derivada solution of each solution of the set S. We begin with the famous Pythagorean equation. Theorem 1.1 Let us consider the diophantine equation x 2 + y 2 = z 2 4) where xyz 0. Then, a complete system of solutions to the equation is x = a 2 b 2, y = 2ab, z = a 2 b 2 5) where a and b are arbitrary integers such that xyz 0. Proof. This equation has solutions x, y, z) such that xyz 0. For example x, y, z) = 3, 4, 5). Let us consider then a solution x, y, z) such that xyz 0. We can write Note that a 0. Consequently see 4)) Therefore x, y, z) = C + a, b, C) 6) C + a + b 2 = C 2 7) C = a2 + b 2 2a 8)

3 Diophantine equations. Elementary methods. 431 Substituting 8) into 7) we obtain a2 + b 2 2a + a + b 2 = a2 + b 2 2a 9) If we now multiply both sides of equation 9) by 2a then we obtain the derivada solution 5) a 2 b 2 + 2ab = a 2 b 2 10) of the solution x, y, z). Note, besides, that 10) is an identity. The theorem is proved. Theorem 1.2 Let us consider the diophantine equation x k x 2 = x 2 h+1 11) where h 2, the coefficients k = 2,..., h) are positive integers and some x = 2,..., h) is different of zero. Then a complete system of solutions to the equation is x 1 = a 2 1 k a 2, x = 2a 1 a = 2,..., h), x h+1 = a 2 1 k a 2 12) where the a = 1,..., h) are arbitrary integers such that some x = 2,..., h) is different of zero. Proof. The equation has solutions with is property, since we have the identity see 12)) 2 2 a 2 1 k a 2 + k 2a 1 a = a 2 1 k a 2 13) Let us consider then a solution x 1,..., x h, x h+1 ) with is property. write We can x 1, x 2,..., x h, x h+1 ) = C + a 1, a 2,..., a h, C) 14) Note that C 0 and a 1 0, since in contrary case the property is not fulfilled. Consequently see 11)) C + a 1 + k a 2 = C 2 15)

4 432 Rafael Jakimczuk Therefore 2Ca 1 + a That is k a 2 = 0 Substituting 16) into 15) we obtain a2 1 + h k a 2 + a 1 + 2a 1 C = a2 1 + h k a 2 2a 1 16) k a 2 = a2 1 + h k a 2 2a 1 17) If we now multiply both sides of equation 17) by 2a 1 then we obtain the derivada solution 12) 2 2 a 2 1 k a 2 + k 2a 1 a = a 2 1 k a 2 18) of the solution x 1,..., x h, x h+1 ). The theorem is proved. Theorem 1.3 Let us consider the diophantine equation k x 2 = k h+1 x 2 h+1 19) where h 2 and the coefficients k = 1,..., h) and k h+1 are positive integers. Suppose that this equation has a solution x 1, x 2,..., x h, x h+1 ) = b 1, b 2,..., b h, 0) different of the trivial solution 0, 0,..., 0, 0) and besides gcdb 1, b 2,..., b h, ) = 1. Then a complete system of solutions is x = b k i c 2 i + 2c k i b i c i = 1, 2,..., h1) x h+1 = where the c i i = 1,..., h) are arbitrary integers. k i c 2 i 22)

5 Diophantine equations. Elementary methods. 433 Proof. Let us consider a solution to equation 19) e 1, e 2,..., e h, C ) where the e = 1,..., h) and C are integers and C 0. Suppose that this solution can not be written in the form b 1 C, b 2 C,..., b h C, C) where C is a integer different of zero. Solutions with is property exist, since we have the identity compare with 21) and 22)) We can write k b k i c 2 i + 2c k i b i c i = k h+1 h k i c 2 i 23) e 1, e 2,..., e h, C ) = b 1 C + a 1, b 2 C + a 2,..., b h C + a h, C4) Consequently and C = C 25) a = e b C = c = 1,..., h6) where the c = 1,..., h) are integers. Note that some a is different of zero and consequently some c is different of zero see 26)), since in contrary case we have see 24)) e 1, e 2,..., e h, C ) = b 1 C, b 2 C,..., b h C, C) where C is given by 25). This is impossible, C can not be a rational not integer since gcdb 1, b 2,..., b h, ) = 1 and C can not be a integer by the established property of the solution see above). Substituting 25) and 26) into 24) we obtain e 1, e 2,..., e h, C ) = b1 C + c 1, b 2 C + c 2,..., Substituting this solution into equation 19) we have k b C + c That is use h k b 2 = k h+1 b 2 h+1 see 20))) b h C + c ) h, C = k h+1 C 27) C 2k b c + k c 2 = 0 28)

6 434 Rafael Jakimczuk That is C = h k c 2 h 2k b c 29) since h k c 2 0 and consequently see 28)) h 2k b c 0. Substituting 29) into 27) we obtain b h k i c 2 i k b h + c = k h+1 h+1 2k i b i c i h k i c 2 i h 2k i b i c i 30) If we now multiply both sides of 30) by b 2 h+1 h 2k i b i c i then we obtain the following derivada solution k b k i c 2 i + 2c k i b i c i = k h+1 h k i c 2 i 31) of the solution e 1, e 2,..., e h, C ) see 31), 21) and 22)). Suppose now that e 1, e 2,..., e h, C ) = b 1 C, b 2 C,..., b h C, C) where C is an integer different of zero. Suppose that some b s = 0, then we put into 21) and 22) c s = C and c = 0 s) and obtain la derivada solution x = b k s C 2 = b C k s C) = 1, 2,..., h) x h+1 = k s C 2 = C k s C) Suppose now that b 0 = 1,..., h), then we put into 21) and 22) c 1 = k 2 b 2 C, c 2 = k 1 b 1 C and c = 0 1, 2) and obtain the derivada solution x = b k 1 c k 2 c 2 2) = b k 1 k 2 b 2 + k 2 k 1 b 1 )C 2 = 1, 2,..., h) x h+1 = k 1 c k 2 c 2 2) = k 1 k 2 b 2 + k 2 k 1 b 1 )C 2 If e 1, e 2,..., e h, C ) = 0, 0,..., 0, 0) then we put c = 0 = 1,..., h) into 21) and 22) and obtain the derivada solution 0, 0,..., 0, 0). The theorem is proved. Example 1.4 Let us consider the diophantine equation Legendre s equation) k 1 x k 2 x 2 2 = k 3 x 2 3

7 Diophantine equations. Elementary methods. 435 Suppose that b 1, b 2, b 3 ) is a solution different of 0, 0, 0) and gcdb 1, b 2, b 3 ) = 1. Then a complete system of solutions is x 1 = b 1 k 1 c k 2 c 2 2) + 2c 1 k 1 b 1 c 1 + k 2 b 2 c 2 ) x 2 = b 2 k 1 c k 2 c 2 2) + 2c 2 k 1 b 1 c 1 + k 2 b 2 c 2 ) x 3 = b 3 k 1 c k 2 c 2 2) where c 1, c 2 and c 3 are arbitrary integers. Theorem 1.5 Let us consider the diophantine equation k x r = k h+1 x M+1 h+1 32) where h 2, the coefficients k = 1,..., h) and k h+1 are integers differents of zero and each integer exponent r 2 = 1,..., h) divides the positive integer M. Let us consider the solutions to the equation x 1,..., x h, x h+1 ) 33) where x h+1 0. Then a complete system of solutions to the equation is M 2 r x = k h+1 A M r b = 1, 2,..., h) 34) where x h+1 = k M 1 h+1 A 35) A = and the b are arbitrary integers such that A 0. k i b r i i 36) Proof. We have the identity compare with 34) and 35)) M 2 r r k k h+1 A M r b = k h+1 k M 1 h+1 A) M+1 37) where A = k b r 0 38)

8 436 Rafael Jakimczuk Consequently the diophantine equation 32) has infinite solutions where x h+1 0. Now, we shall prove that if x 1,..., x h, x h+1 ) 39) is a solution with x h+1 0 then there exist integers b = 1,..., h) such that equations 34) and 35) are a derivada solution of the solution 39). Thus, equations 34) and 35) with the condition A 0 are a complete system of solutions to the equation. Therefore, let us consider a solution to the equation where C 0. Therefore we have x 1,..., x h, x h+1 ) = b 1,..., b h, C) 40) If now we multiply both sides of 41) by k b r = k h+1 C M+1 41) ) k M M+1 ) h+1 C M M+1 ) = k M+1 M ) h+1 C M+1 M then we obtain the derivada solution k M k h+1 C M+1) M r M r r k h+1 ) b = k h+1 k M h+1 C M+1) M+1 42) Equation 41) gives see 42)) x h+1 = k M h+1c M+1 = k M 1 h+1 Therefore see 43) and 42)) we have k b r = k M 1 h+1 A 43) x = kh+1 M 1 M M A) r r k h+1 b = kh+1 A M r b = 1, 2,..., h) 44) M 2 r Consequently equations 43) and 44) are a derivada solution of the solution 40). The theorem is proved. Example 1.6 Now, we give some examples of Theorem 1.5. The equations in a), b) and c) are consider in [2]. a) The equation x 2 + 3y 2 = z 3. x = aa 2 + 3b 2 ), y = ba 2 + 3b 2 ), z = a 2 + 3b 2

9 Diophantine equations. Elementary methods. 437 where a and b are arbitrary integers. b) The equation x 2 + 3y 2 = 4z 3. x = 16aa 2 + 3b 2 ), y = 16ba 2 + 3b 2 ), z = 4a 2 + 3b 2 ) where a and b are arbitrary integers. c) The equation x 4 + 3y 4 = z 5. x = aa 4 + 3b 4 ), y = ba 4 + 3b 4 ), z = a 4 + 3b 4 where a and b are arbitrary integers. d) The equation 3x 3 2y 4 + z 5 = w 121 x = a3a 3 2b 4 + c 5 ) 40, y = b3a 3 2b 4 + c 5 ) 30 z = c3a 3 2b 4 + c 5 4 w = 3a 3 2b 4 + c 5 where a, b and c are arbitrary integers. e) The equation 4x 2 + 3y 4 + 2z 8 = w 9 x = a4a 2 + 3b 4 + 2c 8 ) 4, y = b4a 2 + 3b 4 + 2c 8 z = c4a 2 + 3b 4 + 2c 8 ) w = 4a 2 + 3b 4 + 2c 8 where a, b and c are arbitrary integers. f) The equation x n 1 + x n x n h = x n+1 h+1, x h+1 0) x = b b n i ) = 1,..., h), x h+1 = b n i 45) where b 1, b 2,..., b h are arbitrary integers. In particular, the equation x n +y n = z n+1 has the following complete system of solutions. x = aa n + b n ), y = ba n + b n ), z = a n + b n 46) where a and b are arbitrary integers. Now, we generalize the former theorem.

10 438 Rafael Jakimczuk Theorem 1.7 Let us consider the diophantine equation k x r = k h+1 x M+1 d h+1 where h 2, the coefficients k = 1,..., h) and k h+1 are integers differents of zero, each integer exponent r 2 = 1,..., h) divides the positive integer M and the positive integer d 0 < d < M +1) divides M +1. Let us consider the solutions to the equation x 1,..., x h, x h+1 ) where x h+1 0. Then a complete system of solutions to the equation is M 2 r x = k h+1 A M r b = 1, 2,..., h) x h+1 = kh+1 M 1 A) d where A = h k i b r i i and the b are arbitrary integers such that A 0. Proof. The proof is the same as the former theorem. The theorem is proved. Remark 1.8 Note that the former theorem is a particular case of this theorem when d = 1. Remark 1.9 Note that M +1)/d can be any exponent relatively prime with L, where L is the least common multiple of the exponents r = 1,..., h). Since the linear diophantine equation M+1 y d 1 Ly 2 = 1 have infinite solutions with y 1 > 0 and y 2 > 0, we take M = Ly 2. Example 1.10 Now, we give an example of Theorem 1.7. Let us consider the diophantine equation h k x r = k h+1 x 2 h+1 where the exponents r = 1,..., h) are odd. If we take M = lcmr 1, r 2,..., r h ) then a complete system of solutions to the ecuation is M 2 r x = k h+1 A M r b = 1, 2,..., h) x h+1 = k M 1 h+1 M+1 2 A) where A = h k i b r i i, and the b are arbitrary integers such that A 0. Particular cases of this example appear in [2], for example, x 3 + y 3 = z 2, x 3 + y 3 = 2z 2, x 3 2y 3 = z 2 and another. Acknowledgements. The author is very grateful to Universidad Nacional de Luán. References [1] H. Cohen, Number Theory, Volume I, Springer, [2] H. Cohen, Number Theory, Volume II, Springer, Received: March 8, 2017; Published: March 21, 2017