a;b;c∈N (a+b)(a+2b)(a+3b)=105^c


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Question Number 139051 by mathsuji last updated on 21/Apr/21

$a; b; c \in N$ $(a + b) (a + 2 b) (a + 3 b) = 105^{c}$
Commented by Rasheed.Sindhi last updated on 22/Apr/21

$(a + b) (a + 2 b) (a + 3 b) = 105^{c}$ $(a + b) (a + 2 b) (a + 3 b) = 3^{c} . 5^{c} . 7^{c}$ $a + b = 3^{c} \land a + 2 b = 5^{c} \land a + 3 b = 7^{c}$ $I f c = 1 :$ $a = 1, b = 2, c = 1$ $. . . . .$ $. . .$
Commented by mathsuji last updated on 22/Apr/21

$S i r, s o i t i n f i n i t e o r h o w ?$ $F u l s o l u t i o n p l e a s e I {d i d n}^{'} t$ $u n d e r s t a n d S i r$
Commented by Rasheed.Sindhi last updated on 22/Apr/21

$S i r {i t}^{'} s o n l y a t r y . . . N o t f u l l s o l u t i o n .$
Commented by mathdanisur last updated on 24/Apr/21

$I s t h e r e a c o m p l e t e s o l u t i o n S i r . . .$
Commented by Rasheed.Sindhi last updated on 25/Apr/21

$P l s e e t h e a n s w e r .$
	Answered by Rasheed.Sindhi last updated on 25/Apr/21
	$(a+b)(a+2b)(a+3b)=105^c a,b,c∈N (In the following by N I mean the set: {0,1,2,3,...}) (a+b)(a+2b)(a+3b)=3^c .5^c .7^c Case1:a+b=3^c ,a+2b=5^c ,a+3b=7^c a=3^c −b, 3^c +b=5^c ,3^c +2b=7^c b=5^c −3^c =(1/2)(7^c −3^c ) 2.5^c −2.3^c =7^c −3^c 3^c −2.5^c +7^c =0 c=0,1 (a,b)=(2.3^c −5^c ,5^c −3^c ) (a,b,c)=(1,0,0) (a,b,c)=(1,2,1) Case2:a+b=3^c ,a+2b=7^c ,a+3b=5^c a=3^c −b, 3^c +b=7^c ,3^c +2b=5^c b=7^c −3^c =(1/2){5^c −3^c } 2.7^c −2.3^c =5^c −3^c 3^c +5^c −2.7^c =0 c=0 b=7^c −3^c =7^0 −3^0 =0 a=3^c −(7^c −3^c )=2.3^c −7^c =1 (a,b,c)=(1,0,0) Case3:a+b=5^c ,a+2b=3^c ,a+3b=7^c a=5^c −b, 5^c +b=3^c ,5^c +2b=7^c b=3^c −5^c =(1/2){7^c −5^c } c>0 ⇒b<0 ∴ c=0 b=3^c −5^c =3^0 −5^0 =0 a=5^c −b=5^0 −0=1 (a,b,c)=(1,0,0) Case4:a+b=5^c ,a+2b=7^c ,a+3b=3^c a=5^c −b,5^c +b=7^c ,5^c +2b=3^c b=7^c −5^c =(1/2){3^c −5^c } 2.7^c −2.5^c =3^c −5^c 3^c +5^c −2.7^c =0 c=0 b=7^c −5^c =7^0 −5^0 =0 a=5^c −b=5^0 −0=1 (a,b,c)=(1,0,0) Case5:a+b=7^c ,a+2b=3^c ,a+3b=5^c a=7^c −b, 7^c +b=3^c , 7^c +2b=5^c b=3^c −7^c =(1/2){5^c −7^c ) (Except c=0 b will be negative) c=0 b=3^c −7^c =3^0 −7^0 =0 a=7^c −b=7^0 −0=1 (a,b,c)=(1,0,0) Case6:a+b=7^c ,a+2b=5^c ,a+3b=3^c a=7^c −b , 7^c +b=5^c ,7^c +2b=3^c b=5^c −7^c =(1/2){3^c −7^c } c>0⇒b<0 ∴ c=0 b=5^c −7^c =5^0 −7^0 =0 a=7^c −b=7^0 −0=1 (a,b,c)=(1,0,0) In all cases: (a,b,c)=(1,0,0) or (a,b,c)=(1,2,1) If by N you mean {1,2,3,...} then only one solution: (a,b,c)=(1,2,1)$
	$(a + b) (a + 2 b) (a + 3 b) = 105^{c}$ $a, b, c \in N (I n t h e f o l l o w i n g b y N I$ $m e a n t h e s e t : {0, 1, 2, 3, . . .})$ $(a + b) (a + 2 b) (a + 3 b) = 3^{c} . 5^{c} . 7^{c}$ $C a s e 1 : a + b = 3^{c}, a + 2 b = 5^{c}, a + 3 b = 7^{c}$ $a = 3^{c} - b, 3^{c} + b = 5^{c}, 3^{c} + 2 b = 7^{c}$ $b = 5^{c} - 3^{c} = (1 / 2) (7^{c} - 3^{c})$ $2 . 5^{c} - 2 . 3^{c} = 7^{c} - 3^{c}$ $3^{c} - 2 . 5^{c} + 7^{c} = 0$ $c = 0, 1$ $(a, b) = (2 . 3^{c} - 5^{c}, 5^{c} - 3^{c})$ $(a, b, c) = (1, 0, 0)$ $(a, b, c) = (1, 2, 1)$ $C a s e 2 : a + b = 3^{c}, a + 2 b = 7^{c}, a + 3 b = 5^{c}$ $a = 3^{c} - b, 3^{c} + b = 7^{c}, 3^{c} + 2 b = 5^{c}$ $b = 7^{c} - 3^{c} = (1 / 2) {5^{c} - 3^{c}}$ $2 . 7^{c} - 2 . 3^{c} = 5^{c} - 3^{c}$ $3^{c} + 5^{c} - 2 . 7^{c} = 0$ $c = 0$ $b = 7^{c} - 3^{c} = 7^{0} - 3^{0} = 0$ $a = 3^{c} - (7^{c} - 3^{c}) = 2 . 3^{c} - 7^{c} = 1$ $(a, b, c) = (1, 0, 0)$ $C a s e 3 : a + b = 5^{c}, a + 2 b = 3^{c}, a + 3 b = 7^{c}$ $a = 5^{c} - b, 5^{c} + b = 3^{c}, 5^{c} + 2 b = 7^{c}$ $b = 3^{c} - 5^{c} = (1 / 2) {7^{c} - 5^{c}}$ $c > 0 \Rightarrow b < 0$ $∴ c = 0$ $b = 3^{c} - 5^{c} = 3^{0} - 5^{0} = 0$ $a = 5^{c} - b = 5^{0} - 0 = 1$ $(a, b, c) = (1, 0, 0)$ $C a s e 4 : a + b = 5^{c}, a + 2 b = 7^{c}, a + 3 b = 3^{c}$ $a = 5^{c} - b, 5^{c} + b = 7^{c}, 5^{c} + 2 b = 3^{c}$ $b = 7^{c} - 5^{c} = (1 / 2) {3^{c} - 5^{c}}$ $2 . 7^{c} - 2 . 5^{c} = 3^{c} - 5^{c}$ $3^{c} + 5^{c} - 2 . 7^{c} = 0$ $c = 0$ $b = 7^{c} - 5^{c} = 7^{0} - 5^{0} = 0$ $a = 5^{c} - b = 5^{0} - 0 = 1$ $(a, b, c) = (1, 0, 0)$ $C a s e 5 : a + b = 7^{c}, a + 2 b = 3^{c}, a + 3 b = 5^{c}$ $a = 7^{c} - b, 7^{c} + b = 3^{c}, 7^{c} + 2 b = 5^{c}$ $b = 3^{c} - 7^{c} = (1 / 2) {5^{c} - 7^{c})$ $(E x c e p t c = 0 b w i l l b e n e g a t i v e)$ $c = 0$ $b = 3^{c} - 7^{c} = 3^{0} - 7^{0} = 0$ $a = 7^{c} - b = 7^{0} - 0 = 1$ $(a, b, c) = (1, 0, 0)$ $C a s e 6 : a + b = 7^{c}, a + 2 b = 5^{c}, a + 3 b = 3^{c}$ $a = 7^{c} - b, 7^{c} + b = 5^{c}, 7^{c} + 2 b = 3^{c}$ $b = 5^{c} - 7^{c} = (1 / 2) {3^{c} - 7^{c}}$ $c > 0 \Rightarrow b < 0$ $∴ c = 0$ $b = 5^{c} - 7^{c} = 5^{0} - 7^{0} = 0$ $a = 7^{c} - b = 7^{0} - 0 = 1$ $(a, b, c) = (1, 0, 0)$ $In all cases :$ $(a, b, c) = (1, 0, 0) or (a, b, c) = (1, 2, 1)$ $I f b y N y o u m e a n {1, 2, 3, . . .} t h e n$ $o n l y o n e s o l u t i o n :$ $(a, b, c) = (1, 2, 1)$
	Commented by mathsuji last updated on 28/Apr/21

	$c o o l t h a n k s s i r$
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