XÐT SÈ D− HAI VÕ.VÝ DÔ 5
1. XÐt sè d− hai vÕ.
VÝ dô 5: T×m nghiÖm nguyªn cña ph−¬ng tr×nh:
9x+ =2 y2
+y(*)
Gi¶i:
Ta cã:
VT =9x+ ≡2 2 mod 3( )
⇒VP= y2
+ ≡y 2 mod 3( )
⇔ y y(
+ ≡1) (
2 mod 3)
( )
1 mod 3⇒ y≡( v× nÕu y=3k hoÆc y = 3k+2 th×
VP≡0 mod 3( ) ).
⇒ = +(trong ®ã k
∈Z) thay vµo pt(*) ta cã :
3 1y k( ) (
2
)
2
2
9x+ =2 3k+1 + 3k+ ⇔1 9x=9k +9k⇔ =x k +k = +2
x k k = +VËy
∈k ZVÝ dô 6: Gi¶i ph−¬ng tr×nh nghiÖm nguyªn kh«ng ©m sau:
(
2x
+1 2)(
x
+2 2)(
x
+3 2)(
x
+ −4)
5y
=11879Ta cã
2 ; 2x
x
+1; 2x
+2; 2x
+3; 2x
+4lµ 5 sè tù nhiªn liªn tiÕp
2x
(
2x
+1 2)(
x
+2 2)(
x
+3 2)(
x
+4 5)
⋮MÆt kh¸c ¦CLN(
2x
;5) = 1 nªn (
2x
+1 2)(
x
+2 2)(
x
+3 2)(
x
+4 5)
⋮Víi
y≥1th×
VT =(
2x
+1 2)(
x
+2 2)(
x
+3 2)(
x
+ −4)
5 5y
⋮cßn
VP=11879≡4 mod 5( ) suy ra ph−¬ng
tr×nh kh«ng cã nghiÖm.
Víi y =0 ta cã :
(
2x
+1 2)(
x
+2 2)(
x
+3 2)(
x
+ − =4)
50
11879⇔(
2x
+1 2)(
x
+2 2)(
x
+3 2)(
x
+ =4)
11880(
2x
1 2)(
x
2 2)(
x
3 2)(
x
4)
9.10.11.12 2x
1 9 2x
8 2x
23
x 3⇔ + + + + = ⇒ + = ⇔ = ⇔ = ⇔ =VËy ph−¬ng tr×nh ®; cho cã nghiÖm duy nhÊt ( ) ( )
x y; = 3; 0VÝ dô 7: T×m x, y nguyªn d−¬ng tho¶ m;n :
3x
+ =1(
y+1)
2
( )
2
( )
3x
+ =1 y+1 ⇔3x
= y y+2(**)
Ta cã VT = 3
x
≡ 1 mod 2 ( ) ⇒ VP = y y ( + ≡ 2 ) ( 1 mod 2 )
Suy ra y lµ sè lÎ mµ y vµ y+2 lµ hai sè lÎ liªn tiÕp
=m
3y⇒ + =Tõ pt(**)
n
2 3