Mar 21, 2017 - 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) ...
International Mathematical Forum, Vol. 12, 2017, no. 9, 429 - 438 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/imf.2017.7223
Diophantine Equations. Elementary Methods Rafael Jakimczuk Divisi´on Matem´atica, Universidad Nacional de Luj´an Buenos Aires, Argentina c 2017 Rafael Jakimczuk. This article is distributed under the Creative Copyright 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 P r rh+1 . Some of these diophantine equations are form hj=1 kj xj j = kh+1 xh+1 well-known and our methods of solution are different and very elementary.
Mathematics Subject 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 h X
r
r
h+1 kj xj j = kh+1 xh+1
(1)
j=1
where h ≥ 2, the coefficients kj (j = 1, . . . , h + 1) are integers different of zero and the exponents rj ≥ 2 (j = 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 (x1 , x2 , . . . , xh , xh+1 )
(2)
to equation (1) (the xj (j = 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
430
Rafael Jakimczuk
the least common multiple (lcm)of the exponents rj (j = 1, . . . , h + 1), then we obtain the solution
L rh
L r2
L r1
x1 E , x2 E , . . . , xh E , xh+1 E
L rh+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 x2 + 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 x2 + y 2 = z 2
(4)
where xyz 6= 0. Then, a complete system of solutions to the equation is x = a2 − b 2 ,
y = 2ab,
z = −a2 − b2
(5)
where a and b are arbitrary integers such that xyz 6= 0. Proof. This equation has solutions (x, y, z) such that xyz 6= 0. For example (x, y, z) = (3, 4, 5). Let us consider then a solution (x, y, z) such that xyz 6= 0. We can write (x, y, z) = (C + a, b, C)
(6)
Note that a 6= 0. Consequently (see (4)) (C + a)2 + b2 = C 2
(7)
Therefore C=−
a2 + b 2 2a
(8)
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Diophantine equations. Elementary methods.
Substituting (8) into (7) we obtain a2 + b 2 +a − 2a
!2
a2 + b 2 +b = − 2a
!2
2
(9)
If we now multiply both sides of equation (9) by (2a)2 then we obtain the derivada solution (5)
a2 − b 2
2
+ (2ab)2 = −a2 − b2
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 x21
+
h X
kj x2j = x2h+1
(11)
j=2
where h ≥ 2, the coefficients kj (j = 2, . . . , h) are positive integers and some xj (j = 2, . . . , h) is different of zero. Then a complete system of solutions to the equation is x1 = a21 −
h X
kj a2j ,
xj = 2a1 aj
(j = 2, . . . , h),
xh+1 = −a21 −
h X
kj a2j (12)
j=2
j=2
where the aj (j = 1, . . . , h) are arbitrary integers such that some xj (j = 2, . . . , h) is different of zero. Proof. The equation has solutions with is property, since we have the identity (see (12)) a2 1
−
h X j=2
2
kj a2j +
h X
kj (2a1 aj )2 = −a21 −
h X
2
kj a2j
(13)
j=2
j=2
Let us consider then a solution (x1 , . . . , xh , xh+1 ) with is property. We can write (x1 , x2 , . . . , xh , xh+1 ) = (C + a1 , a2 , . . . , ah , C)
(14)
Note that C 6= 0 and a1 6= 0 , since in contrary case the property is not fulfilled. Consequently (see (11)) (C + a1 )2 +
h X j=2
kj a2j = C 2
(15)
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Rafael Jakimczuk
Therefore 2Ca1 + a21 +
h X
kj a2j = 0
j=2
That is C=−
a21 +
Ph
j=2
kj a2j
(16)
2a1
Substituting (16) into (15) we obtain −
a21 +
Ph
j=2
kj a2j
2a1
!2
+ a1
+
h X
kj a2j
= −
a21 +
j=2
Ph
j=2
kj a2j
2a1
!2
(17)
If we now multiply both sides of equation (17) by (2a1 )2 then we obtain the derivada solution (12) a2 1
−
h X
2
kj a2j +
j=2
h X
2
kj (2a1 aj ) = −a21 −
h X
2
kj a2j
(18)
j=2
j=2
of the solution (x1 , . . . , xh , xh+1 ). The theorem is proved. Theorem 1.3 Let us consider the diophantine equation h X
kj x2j = kh+1 x2h+1
(19)
j=1
where h ≥ 2 and the coefficients kj (j = 1, . . . , h) and kh+1 are positive integers. Suppose that this equation has a solution (x1 , x2 , . . . , xh , xh+1 ) = (b1 , b2 , . . . , bh , bh+1 )
(20)
different of the trivial solution (0, 0, . . . , 0, 0) and besides gcd(b1 , b2 , . . . , bh , bh+1 ) = 1. Then a complete system of solutions is xj = −bj
h X i=1
ki c2i + 2cj
h X
ki bi ci
(j = 1, 2, . . . , h)
(21)
i=1
xh+1 = −bh+1
h X
ki c2i
i=1
where the ci (i = 1, . . . , h) are arbitrary integers.
(22)
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Diophantine equations. Elementary methods.
Proof. Let us consider a solution to equation (19) (e1 , e2 , . . . , eh , C 0 ) where the ej (j = 1, . . . , h) and C 0 are integers and C 0 6= 0. Suppose that this solution can not be written in the form (b1 C, b2 C, . . . , bh C, bh+1 C) where C is a integer different of zero. Solutions with is property exist, since we have the identity (compare with (21) and (22)) h X
kj −bj
j=1
h X
ki c2i
+ 2cj
i=1
h X
!2
= kh+1 −bh+1
ki bi ci
i=1
h X
!2
ki c2i
(23)
i=1
We can write (e1 , e2 , . . . , eh , C 0 ) = (b1 C + a1 , b2 C + a2 , . . . , bh C + ah , bh+1 C)
(24)
Consequently C=
C0 bh+1
(25)
and aj = e j −
bj bh+1
C0
=
cj
(j = 1, . . . , h)
bh+1
(26)
where the cj (j = 1, . . . , h) are integers. Note that some aj is different of zero and consequently some cj is different of zero (see (26)), since in contrary case we have (see (24)) (e1 , e2 , . . . , eh , C 0 ) = (b1 C, b2 C, . . . , bh C, bh+1 C) where C is given by (25). This is impossible, C can not be a rational not integer since gcd(b1 , b2 , . . . , bh , bh+1 ) = 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 0
(e1 , e2 , . . . , eh , C ) =
b1
0
bh+1
C +
c1
b2
0
c2
bh
0
ch
!
, C + ,..., C + ,C bh+1 bh+1 bh+1 bh+1 bh+1
0
Substituting this solution into equation (19) we have h X j=1
That is (use
Ph
j=1
kj
bj
0
bh+1
C +
cj
!2
bh+1
2
= kh+1 (C 0 )
(27)
kj b2j = kh+1 b2h+1 (see (20)))
C0
h X
j=1
2kj bj cj +
h X j=1
kj c2j = 0
(28)
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Rafael Jakimczuk
That is 0
Ph
kj c2j j=1 2kj bj cj
−
C = Ph
j=1
since hj=1 kj c2j 6= 0 and consequently (see (28)) Substituting (29) into (27) we obtain P
h X j=1
kj
Ph
ki c2i cj + Ph bh+1 i=1 2ki bi ci bh+1 bj −
(29) Ph
j=1
!2
i=1
= kh+1
h X j=1
kj −bj
h X i=1
ki c2i
+ 2cj
h X
Ph
ki c2i Ph i=1 2ki bi ci −
Ph
If we now multiply both sides of (30) by b2h+1 ( the following derivada solution
2kj bj cj 6= 0.
i=1
(30)
2ki bi ci )2 then we obtain
!2
ki bi ci
!2
i=1
= kh+1 −bh+1
h X
!2
ki c2i
(31)
i=1
i=1
of the solution (e1 , e2 , . . . , eh , C 0 ) (see (31), (21) and (22)). Suppose now that (e1 , e2 , . . . , eh , C 0 ) = (b1 C, b2 C, . . . , bh C, bh+1 C) where C is an integer different of zero. Suppose that some bs = 0, then we put into (21) and (22) cs = C and cj = 0 (j 6= s) and obtain la derivada solution xj = −bj ks C 2 = bj C(−ks C)
(j = 1, 2, . . . , h)
xh+1 = −bh+1 ks C 2 = bh+1 C(−ks C) Suppose now that bj 6= 0 (j = 1, . . . , h), then we put into (21) and (22) c1 = −k2 b2 C, c2 = k1 b1 C and cj = 0 (j 6= 1, 2) and obtain the derivada solution xj = −bj (k1 c21 + k2 c22 ) = −bj (k1 (−k2 b2 )2 + k2 (k1 b1 )2 )C 2
(j = 1, 2, . . . , h)
xh+1 = −bh+1 (k1 c21 + k2 c22 ) = −bh+1 (k1 (−k2 b2 )2 + k2 (k1 b1 )2 )C 2 If (e1 , e2 , . . . , eh , C 0 ) = (0, 0, . . . , 0, 0) then we put cj = 0 (j = 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) k1 x21 + k2 x22 = k3 x23
435
Diophantine equations. Elementary methods.
Suppose that (b1 , b2 , b3 ) is a solution different of (0, 0, 0) and gcd(b1 , b2 , b3 ) = 1. Then a complete system of solutions is x1 = −b1 (k1 c21 + k2 c22 ) + 2c1 (k1 b1 c1 + k2 b2 c2 ) x2 = −b2 (k1 c21 + k2 c22 ) + 2c2 (k1 b1 c1 + k2 b2 c2 ) x3 = −b3 (k1 c21 + k2 c22 ) where c1 , c2 and c3 are arbitrary integers. Theorem 1.5 Let us consider the diophantine equation h X
r
+1 kj xj j = kh+1 xM h+1
(32)
j=1
where h ≥ 2, the coefficients kj (j = 1, . . . , h) and kh+1 are integers differents of zero and each integer exponent rj ≥ 2 (j = 1, . . . , h) divides the positive integer M . Let us consider the solutions to the equation (x1 , . . . , xh , xh+1 )
(33)
where xh+1 6= 0. Then a complete system of solutions to the equation is M2 rj
M
xj = kh+1 A rj bj
(j = 1, 2, . . . , h)
M −1 xh+1 = kh+1 A
(34) (35)
where A=
h X
ki bri i
(36)
i=1
and the bj are arbitrary integers such that A 6= 0. Proof. We have the identity (compare with (34) and (35)) h X
M2 rj
M rj
rj
M +1
M −1 kj kh+1 A bj = kh+1 kh+1 A
(37)
j=1
where A=
h X j=1
r
kj bj j 6= 0
(38)
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Rafael Jakimczuk
Consequently the diophantine equation (32) has infinite solutions where xh+1 6= 0. Now, we shall prove that if (x1 , . . . , xh , xh+1 )
(39)
is a solution with xh+1 6= 0 then there exist integers bj (j = 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 6= 0 are a complete system of solutions to the equation. Therefore, let us consider a solution to the equation (x1 , . . . , xh , xh+1 ) = (b1 , . . . , bh , C)
(40)
where C 6= 0. Therefore we have h X
r
kj bj j = kh+1 C M +1
(41)
j=1
If now we multiply both sides of (41) by
M kh+1
M +1
CM
M +1
M +1 = kh+1
M
C M +1
M
then we obtain the derivada solution h X
kj
M kh+1 C M +1
M rj
M rj
!rj
M = kh+1 kh+1 C M +1
kh+1 bj
M +1
(42)
j=1
Equation (41) gives (see (42)) M −1 M xh+1 = kh+1 C M +1 = kh+1
h X
r kj bj j
M −1 = kh+1 A
(43)
j=1
Therefore (see (43) and (42)) we have xj =
M −1 kh+1 A
M rj
M rj
M2 rj
M
kh+1 bj = kh+1 A rj bj
(j = 1, 2, . . . , h)
(44)
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 x2 + 3y 2 = z 3 . This equation has the following complete system of solutions. x = a(a2 + 3b2 ),
y = b(a2 + 3b2 ),
z = a2 + 3b2
437
Diophantine equations. Elementary methods.
where a and b are arbitrary integers. b) The equation x2 + 3y 2 = 4z 3 . This equation has the following complete system of solutions. x = 16a(a2 + 3b2 ),
y = 16b(a2 + 3b2 ),
z = 4(a2 + 3b2 )
where a and b are arbitrary integers. c) The equation x4 + 3y 4 = z 5 . This equation has the following complete system of solutions. x = a(a4 + 3b4 ),
y = b(a4 + 3b4 ),
z = a4 + 3b4
where a and b are arbitrary integers. d) The equation 3x3 − 2y 4 + z 5 = w121 This equation has the following complete system of solutions. x = a(3a3 − 2b4 + c5 )40 , z = c(3a3 − 2b4 + c5 )24
y = b(3a3 − 2b4 + c5 )30 w = 3a3 − 2b4 + c5
where a, b and c are arbitrary integers. e) The equation 4x2 + 3y 4 + 2z 8 = w9 This equation has the following complete system of solutions. x = a(4a2 + 3b4 + 2c8 )4 , z = c(4a2 + 3b4 + 2c8 )
y = b(4a2 + 3b4 + 2c8 )2 w = 4a2 + 3b4 + 2c8
where a, b and c are arbitrary integers. f ) The equation xn1 + xn2 + · · · + xnh = xn+1 h+1 , (xh+1 6= 0) This equation has the following complete system of solutions. xj = b j
h X
!
bni
(j = 1, . . . , h),
i=1
xh+1 =
h X
bni
(45)
i=1
where b1 , b2 , . . . , bh are arbitrary integers. In particular, the equation xn +y n = z n+1 has the following complete system of solutions. x = a(an + bn ),
y = b(an + bn ),
where a and b are arbitrary integers. Now, we generalize the former theorem.
z = an + b n
(46)
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Rafael Jakimczuk
Theorem 1.7 Let us consider the diophantine equation h X
r
M +1
d kj xj j = kh+1 xh+1
j=1
where h ≥ 2, the coefficients kj (j = 1, . . . , h) and kh+1 are integers differents of zero, each integer exponent rj ≥ 2 (j = 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 (x1 , . . . , xh , xh+1 ) where xh+1 6= 0. Then a complete system of solutions to the equation is M2 rj
M
xj = kh+1 A rj bj where A =
Ph
i=1
(j = 1, 2, . . . , h)
M −1 xh+1 = kh+1 A
d
ki bri i and the bj are arbitrary integers such that A 6= 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 rj (j = 1, . . . , h). Since the linear diophantine equation Md+1 y1 − Ly2 = 1 have infinite solutions with y1 > 0 and y2 > 0, we take M = Ly2 . Example 1.10 Now, we give an example of Theorem 1.7. Let us consider P r the diophantine equation hj=1 kj xj j = kh+1 x2h+1 where the exponents rj (j = 1, . . . , h) are odd. If we take M = lcm(r1 , r2 , . . . , rh ) then a complete system of solutions to the ecuation is M2 rj
M
xj = kh+1 A rj bj
(j = 1, 2, . . . , h)
M +1
M −1 xh+1 = kh+1 A
2
where A = hi=1 ki bri i , and the bj are arbitrary integers such that A 6= 0. Particular cases of this example appear in [2], for example, x3 + y 3 = z 2 , x3 + y 3 = 2z 2 , x3 − 2y 3 = z 2 and another. P
Acknowledgements. The author is very grateful to Universidad Nacional de Luj´an.
References [1] H. Cohen, Number Theory, Volume I, Springer, 2010. [2] H. Cohen, Number Theory, Volume II, Springer, 2010. Received: March 8, 2017; Published: March 21, 2017
