Bézouten identitatea k (edo Bézouten lema k) zera dio: zeroren ezberdinak diren
a
{\displaystyle a}
b
{\displaystyle b}
zenbaki oso elkarrekiko lehenak badira, orduan existitzen dira x eta y bi zenbaki oso non 1=
a
x
+
b
y
=
d
{\displaystyle ax+by=d}
Era berean,
d
=
z
k
h
(
a
,
b
)
{\displaystyle d=zkh(a,b)}
Alde batetik, baldin eta
n
≠
0
{\displaystyle n\neq 0}
z
k
h
(
n
a
,
n
b
)
=
|
n
|
z
k
h
(
a
,
b
)
{\displaystyle zkh(na,nb)=|n|zkh(a,b)}
Bestalde,
z
k
h
(
a
/
c
,
b
/
d
)
=
1
{\displaystyle zkh(a/c,b/d)=1}
Identitateari izena Étienne Bézout (1730-1783) matematikari frantsesaren omenez jarri zitzaion[1] Zenbaki teoriako beste teorema batzuk (Euklidesen lema edo Hondarraren teorema txinatarra , adibidez) Bézouten identitatean oinarritzen dira.
Soluzioen egitura
aldatu
Bézouten identitateko (
x
{\displaystyle x}
y
{\displaystyle y}
a
{\displaystyle a}
b
{\displaystyle b}
d
=
z
k
h
(
a
,
b
)
{\displaystyle d=zkh(a,b)}
zatitzaile komun handiena Euklidesen algoritmo hedatuaren bitartez kalkula daitezke eta lortuko diren balioek honakoa beteko dute:
|
x
|
≤
|
b
d
|
{\displaystyle |x|\leq \left|{\frac {b}{d}}\right|}
|
y
|
≤
|
a
d
|
{\displaystyle |y|\leq \left|{\frac {a}{d}}\right|}
a
{\displaystyle a}
b
{\displaystyle b}
Bézouten koefizienteak ez dira bakarrak. Izan ere,
a
(
x
−
k
b
)
+
b
(
y
+
k
a
)
=
a
x
−
k
b
a
+
b
y
+
k
b
a
=
a
x
+
b
y
{\displaystyle a(x-kb)+b(y+ka)=ax-kba+by+kba=ax+by}
Hortaz, (
x
{\displaystyle x}
y
{\displaystyle y}
Euklidesen algoritmo hedatuaren bidez, adibidez), gainerako (
x
′
{\displaystyle x'}
y
′
{\displaystyle y'}
x
′
=
x
−
k
b
,
y
′
=
y
+
k
a
{\displaystyle x^{\prime }=x-kb,\qquad y^{\prime }=y+ka}
Izan bitez
a
=
502
{\displaystyle a=502}
b
=
110
{\displaystyle b=110}
zenbaki oso . Euklidesen algoritmo hedatuaren bidez
(
x
,
y
)
{\displaystyle (x,y)}
zatitzaile komun handiena den
d
=
z
k
h
(
a
,
b
)
{\displaystyle d=zkh(a,b)}
Zatiketa euklidearrak eginek eta hondarrak askatuz, zera lortuko dugu:
502
=
110
(
4
)
+
62
⇒
62
=
502
(
1
)
−
110
(
4
)
⇒
62
=
502
(
1
)
+
110
(
−
4
)
110
=
62
(
1
)
+
48
⇒
48
=
110
(
1
)
−
62
(
1
)
⇒
48
=
110
(
1
)
+
62
(
−
1
)
62
=
48
(
1
)
+
14
⇒
14
=
62
(
1
)
−
48
(
1
)
⇒
14
=
62
(
1
)
+
48
(
−
1
)
48
=
14
(
3
)
+
6
⇒
6
=
48
(
1
)
−
14
(
3
)
⇒
6
=
48
(
1
)
+
14
(
−
3
)
14
=
6
(
2
)
+
2
⇒
2
=
14
(
1
)
−
6
(
2
)
⇒
2
=
14
(
1
)
+
6
(
−
2
)
{\displaystyle {\begin{aligned}&502=110(4)+62\quad \Rightarrow \quad 62=502(1)-110(4)\quad \Rightarrow \quad 62=502(1)+110(-4)\\&110=62(1)+48\quad \Rightarrow \quad 48=110(1)-62(1)\quad \Rightarrow \quad 48=110(1)+62(-1)\\&62=48(1)+14\quad \Rightarrow \quad 14=62(1)-48(1)\quad \Rightarrow \quad 14=62(1)+48(-1)\\&48=14(3)+6\quad \Rightarrow \quad 6=48(1)-14(3)\quad \Rightarrow \quad 6=48(1)+14(-3)\\&14=6(2)+2\quad \Rightarrow \quad 2=14(1)-6(2)\quad \Rightarrow \quad 2=14(1)+6(-2)\end{aligned}}}
z
k
h
(
502
,
110
)
=
2
{\displaystyle zkh(502,110)=2}
x
{\displaystyle x}
y
{\displaystyle y}
2
=
502
(
x
)
+
100
(
y
)
2
=
14
(
1
)
+
6
(
−
2
)
=
[
62
(
1
)
+
48
(
−
1
)
]
(
1
)
+
[
48
(
1
)
+
14
(
−
3
)
]
(
−
2
)
=
[
62
(
1
)
+
48
(
−
1
)
]
+
[
48
(
−
2
)
+
14
(
6
)
]
=
62
(
1
)
+
48
(
−
3
)
+
14
(
6
)
=
[
502
(
1
)
+
110
(
−
4
)
]
(
1
)
+
[
110
(
1
)
+
62
(
−
1
)
]
(
−
3
)
+
[
62
(
1
)
+
48
(
−
1
)
]
(
6
)
=
[
502
(
1
)
+
110
(
−
4
)
]
(
1
)
+
[
110
(
−
3
)
+
62
(
3
)
]
+
[
62
(
6
)
+
48
(
−
6
)
]
=
[
502
(
1
)
+
110
(
−
7
)
+
62
(
9
)
+
48
(
−
6
)
]
=
[
502
(
1
)
+
110
(
−
7
)
]
+
[
502
(
9
)
+
110
(
−
36
)
+
[
110
(
−
6
)
+
62
(
−
6
)
]
]
=
[
502
(
10
)
+
110
(
−
49
)
+
62
(
6
)
]
=
[
502
(
10
)
+
110
(
−
49
)
]
+
[
502
(
1
)
+
110
(
−
4
)
]
(
6
)
=
[
502
(
10
)
+
110
(
−
49
)
]
+
[
502
(
6
)
+
110
(
−
24
)
]
=
502
(
16
)
+
110
(
−
73
)
{\displaystyle {\begin{aligned}2&=502(x)+100(y)\\2&=14(1)+6(-2)\\&=[62(1)+48(-1)](1)+[48(1)+14(-3)](-2)\\&=[62(1)+48(-1)]+[48(-2)+14(6)]\\&=62(1)+48(-3)+14(6)\\&=[502(1)+110(-4)](1)+[110(1)+62(-1)](-3)+[62(1)+48(-1)](6)\\&=[502(1)+110(-4)](1)+[110(-3)+62(3)]+[62(6)+48(-6)]\\&=[502(1)+110(-7)+62(9)+48(-6)]\\&=[502(1)+110(-7)]+[502(9)+110(-36)+[110(-6)+62(-6)]]\\&=[502(10)+110(-49)+62(6)]\\&=[502(10)+110(-49)]+[502(1)+110(-4)](6)\\&=[502(10)+110(-49)]+[502(6)+110(-24)]\\&=502(16)+110(-73)\end{aligned}}}
x
=
16
{\displaystyle x=16}
y
=
−
73
{\displaystyle y=-73}
Egiazta daiteke Bézouten identitatea (
a
x
+
b
y
=
d
{\displaystyle ax+by=d}
502
(
16
)
+
110
(
−
73
)
=
2
{\displaystyle 502(16)+110(-73)=2}
Esan dugunez, Bézouten koefizienteak ez dira bakarrak.
k
=
1
{\displaystyle k=1}
x
′
{\displaystyle x'}
y
′
{\displaystyle y'}
x
′
=
x
−
k
b
=
16
−
110
=
−
94
,
y
′
=
y
+
k
a
=
−
73
+
502
=
429
{\displaystyle x^{\prime }=x-kb=16-110=-94,\,\qquad y^{\prime }=y+ka=-73+502=429}
a
x
+
b
y
=
z
k
h
(
a
,
b
)
{\displaystyle ax+by=zkh(a,b)}
502
(
−
94
)
+
110
(
429
)
=
2
{\displaystyle 502(-94)+110(429)=2}
Erreferentziak
aldatu
Kanpo estekak
aldatu
+[48(1)+14(-3)](-2)\\&=[62(1)+48(-1)]+[48(-2)+14(6)]\\&=62(1)+48(-3)+14(6)\\&=[502(1)+110(-4)](1)+[110(1)+62(-1)](-3)+[62(1)+48(-1)](6)\\&=[502(1)+110(-4)](1)+[110(-3)+62(3)]+[62(6)+48(-6)]\\&=[502(1)+110(-7)+62(9)+48(-6)]\\&=[502(1)+110(-7)]+[502(9)+110(-36)+[110(-6)+62(-6)]]\\&=[502(10)+110(-49)+62(6)]\\&=[502(10)+110(-49)]+[502(1)+110(-4)](6)\\&=[502(10)+110(-49)]+[502(6)+110(-24)]\\&=502(16)+110(-73)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99bc33b18be776a03a019ad94567bd1ca3ea14b9)
Datuak: Q513028