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9Вестник Сыктывкарского университета. Сер.1. Вып.1(20). 2015
УДК 511.0
SUFFICIENT AND NECESSARY CONDITION FOR THE SOLUTION OF THE BEAL CONJECTURE
A. Grytczuk
In this paper we prove some sufficient and necessary condition connected with the Beal conjecture. In 1993 year Beal formulated the following conjecture: if the diophantine equation (*) ax + by = cz has a solution in positive integers a, b, c, x, y, z such that x > 2, y > 2, z > 2 then the numbers a, b, c have a prime common factor. The following result is proved in this paper: The equation (*) has a solution in positive integers a, b, c, x, y, z such that x > 2, y > 2, z > 2 and a, b, c are pairwise relative primes with by > ax if and only if there is some integer r1; 1 < r1 < ax such that (**) by = ax + ri, cz = 2 ■ ax + r1. In the proof of this result we use some properties of the divisibility relation.
Keywords: Diophantine equations, Beal's conjecture.
In this Note we present some result concerning the equation
ax + by = cz (*)
where a, b, c, x, y, z are positive integers such that x > 2, y > 2 and z > 2.
1. Introduction
In 1993 Beal formulated the following conjecture:
If the equation (*) has a solution in positive integers a, b, c, x, y, z such that x> 2, y> 2, z> 2 then the numbers a, b, c have a prime common factor.
In this connection we prove the following Theorem:
Theorem. If the numbers a, b, c, x, y, z are positive integers such that x> 2, y> 2, z> 2 and (a, b) = 1, (b, c) = 1, (a, c) = 1 and the equation (*) has a solution then there is positive integer 1 < r1 < ax such that
by = ax + ri, cz = 2ax + ri. (1.1)
© Grytczuk A., 2015.
2. Basic Lemmas
In the proof of the Theorem we use some well-known results from elementary number theory given in Lemma 1 and Lemma 2.
Lemma 1. ([1] Th.1.3, p.10). Let a, b be positive integers. Then there exist unique integers q, r such that
b = a • q + r, where 0 < r < a. (2.1)
Lemma 2. Fundamental properties of the divisibility relation.
Let a, b, c are some integers numbers. Then we have:
a | b ^ there is q G Z such that b = a • q. (2.2)
If a | b and b | c then a | c. (2.3)
If a | b then for any integer k = 0 we have b = a • k. (2.4)
If c | a and c | b then c | a + b and c | a — b. (2.5)
3.Proof of the Theorem
Suppose that the equation (*) has a solution with integer numbers satisfying the assumption of the Theorem. Then we have
ax < cz and by < cz. (3.1)
Moreover, without loss of generality we can assumed that
ax < by. (3.2)
From Lemma 1 and (3.1),(3.2) and the assumption of the Theorem that a, b, c are pairwise relatively prime we have that (ax,by) = 1, (cz, by) = 1, (cz, ax) = 1, hence we get
by = ax • qi + ri; qi > 1,1 < ri < ax; (3.3)
cz = by • q2 + r2; q2 > 1,1 < r2 < by; (3.4)
cz = ax • qs + rs; qs > 1,1 < rs < ax.
(3.5)
Since (cz, by) = 1, then from Theorem16 on page 25, see [2], it follows that there exist positive integers £, n such that
. £ - by ■ n =1, (£,n) = 1. (3.6)
From (3.4) and (3.5) we get
by ■ q2 + r2 = ax ■ qs + rs. (3.7)
By (3.3) and (3.7) it follows that
ax (qi ■ q2 - qs) + ri ■ q2 + r2 - rs = 0. (3.8)
On the other hand from (*),(3.3) and (3.5) we obtain
ax (1 + qi - qs) = rs - ri. (3.9)
If rs - ri = 0 then by (3.9) and the fact that ax > 1 it follows that 1 + qi - qs = 0. From (3.9) and (2.2) of Lemma 2 it follws that
ax | rs - ri. (3.10)
Since rs - ri = 0 then we have two cases: 10.rs - ri > 0 or 20.rs - ri < 0. In the case 10 by (3.10) we get
ax < rs - ri < rs. (3.11)
From (3.11) and (3.5) it follows that ax < rs < ax, so is impossible. In the case 20 from (3.10) and (2.4) when k = -1, we have that ax | (-1)-(rs - ri), so denote that
ax | ri - rs. (3.12)
By (3.12) it follows that
ax < ri - rs <ri. (3.13)
From (3.13) and (3.3) it follows that ax < ri < ax, so is impossible. Consequently from (3.8) we obtain
rs = ri, qs = qi + 1. (3.14)
Now, from (*) and (3.2) we have
ax + by = by ■ q2 + r2 ^ ax - r2 = by ■ (q2 - 1). (3.15)
If q2 > 1 then from (3.15) we have that ax — r2 > 0 and consequently by (2.2) of Lemma 2 it follows that
by | ax — r2. (3.16)
From (3.16) we get that by < ax — r2 < ax, so contradicts to inequality (3.2). Since q2 > 1 and the inequality q2 > 1 is impossible then by (3.15) it follows that
q2 = 1 and ax = r2. (3.17)
Now, we use the following well-known identity
4 ■ r ■ s = (r + s)2 — (r — s)2 . (**)
Let r = cz, s = by, then by (**), (3.14), (*) and (3.3)-(3.5) it follows that
4 ■ (ax ■ qs + rs) ■ by + a2x = (ax ■ qs + rs)2 + 2 ■ (ax ■ qs + rs) ■ by + b2y. (3.18) From (3.18) we obtain
by ■ (2ax ■ qs + 2rs — by) = ax ■ [ax ■ (q32 — 1) + 2qs ■ rs] + r32. (3.19) Since rs = ri and by = ax ■ qi + ri, then from (3.18) we have
by ■ (2qs — qi) + qi ■ ri = ax ■ (q32 — 1) + 2qs ■ qi. (3.20)
By (3.20) and (3.3) it follows that
ax ■ [q2 — 1 — qi ■ (2qs — 1)] = ri ■ (qi — 1). (3.21)
From (3.14) we have
q2 — 1 — qi ■ (2qs — 1) = qi — q2 = —qi ■ (qi — 1). (3.22) Hence, from (3.22) and (3.21) we get
(qi — 1) ■ (ax ■ qi + ri) = (qi — 1) ■ by = 0. (3.23)
By (3.23), (3.14) and the fact that by > 0 it follows that
qi = 1, qs = qi + 1 = 2. (* * *)
From (* * *), (3.3), (3.5) and (3.14) we obtain that by = ax + ri and cz = 2ax + ri and the proof of the Theorem is complete.
Remark
We note that if the conditions given in the formulas (1.1) are satisfied then the equation (*) has a solution. Realy, we have
cz = 2ax + r1 = ax + ax + r1 = ax + by.
From ths fact and the Theorem we obtain the following sufficient and necessary condition for the solution of the Beal conjecture:
The equation (*) has a solution in positive integers a, b, c, x, y, z such that x > 2, y > 2, z > 2 and (a, b) = 1, (b, c) = 1, (a, c) = 1, by > ax if and only if for some integer r1 with 1 < r < ax the conditions (1.1) are satisfied.
References
1. Redmond D. Number Theory, Mercel Dekker, Inc. New York. Basel.
Hong-Kong, 1996.
2. Sierpinski W. Elementary Number Theory, PWN Warszawa, 1987.
Аннотация
Грытчук A. Достаточные и необходимые условия для решения гипотезы Била
В 1993 году Эндрю „Энди" Бил (Andrew „Andy" Beal) высказал гипотезу: Если (*) ax + by = cz, где a, b, c, x, y, z — положительные целые числа и x, y, z строго больше 2, то a, b и c должны иметь общий простой делитель. В работе получено необходимое и достаточное условие решение уравнения (*) в положительных целых числах a, b, c, x, y, z, таких, что x > 2, y > 2, z > 2 и числа a, b, c попарно взаимно просты и by > ax.
Ключевые слова: гипотеза Била, диофантовы уравнения, простой делитель.
University of Zielona Gora, Poland
Поступила 01.09.2015
