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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∈N (a+b)(a+2b)(a+3b)=105^c
a;b;c∈N(a+b)(a+2b)(a+3b)=105c
Commented by Rasheed.Sindhi last updated on 22/Apr/21
(a+b)(a+2b)(a+3b)=105^c (a+b)(a+2b)(a+3b)=3^c .5^c .7^c a+b=3^c ∧a+2b=5^c ∧ a+3b=7^c If c=1: a=1,b=2,c=1 ..... ...
(a+b)(a+2b)(a+3b)=105c(a+b)(a+2b)(a+3b)=3c.5c.7ca+b=3c∧a+2b=5c∧a+3b=7cIfc=1:a=1,b=2,c=1…..…
Commented by mathsuji last updated on 22/Apr/21
Sir, so it infinite or how? Ful solution please I didn′t understand Sir
Sir,soitinfiniteorhow?FulsolutionpleaseIdidn′tunderstandSir
Commented by Rasheed.Sindhi last updated on 22/Apr/21
Sir it′s only a try...Not full solution.
Sirit′sonlyatry…Notfullsolution.
Commented by mathdanisur last updated on 24/Apr/21
Is there a complete solution Sir...
IsthereacompletesolutionSir…
Commented by Rasheed.Sindhi last updated on 25/Apr/21
Pl see the answer.
Plseetheanswer.
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+2b)(a+3b)=105ca,b,c∈N(InthefollowingbyNImeantheset:{0,1,2,3,…})(a+b)(a+2b)(a+3b)=3c.5c.7cCase1:a+b=3c,a+2b=5c,a+3b=7ca=3c−b,3c+b=5c,3c+2b=7cb=5c−3c=(1/2)(7c−3c)2.5c−2.3c=7c−3c3c−2.5c+7c=0c=0,1(a,b)=(2.3c−5c,5c−3c)(a,b,c)=(1,0,0)(a,b,c)=(1,2,1)Case2:a+b=3c,a+2b=7c,a+3b=5ca=3c−b,3c+b=7c,3c+2b=5cb=7c−3c=(1/2){5c−3c}2.7c−2.3c=5c−3c3c+5c−2.7c=0c=0b=7c−3c=70−30=0a=3c−(7c−3c)=2.3c−7c=1(a,b,c)=(1,0,0)Case3:a+b=5c,a+2b=3c,a+3b=7ca=5c−b,5c+b=3c,5c+2b=7cb=3c−5c=(1/2){7c−5c}c>0⇒b<0∴c=0b=3c−5c=30−50=0a=5c−b=50−0=1(a,b,c)=(1,0,0)Case4:a+b=5c,a+2b=7c,a+3b=3ca=5c−b,5c+b=7c,5c+2b=3cb=7c−5c=(1/2){3c−5c}2.7c−2.5c=3c−5c3c+5c−2.7c=0c=0b=7c−5c=70−50=0a=5c−b=50−0=1(a,b,c)=(1,0,0)Case5:a+b=7c,a+2b=3c,a+3b=5ca=7c−b,7c+b=3c,7c+2b=5cb=3c−7c=(1/2){5c−7c)(Exceptc=0bwillbenegative)c=0b=3c−7c=30−70=0a=7c−b=70−0=1(a,b,c)=(1,0,0)Case6:a+b=7c,a+2b=5c,a+3b=3ca=7c−b,7c+b=5c,7c+2b=3cb=5c−7c=(1/2){3c−7c}c>0⇒b<0∴c=0b=5c−7c=50−70=0a=7c−b=70−0=1(a,b,c)=(1,0,0)Inallcases:(a,b,c)=(1,0,0)or(a,b,c)=(1,2,1)IfbyNyoumean{1,2,3,…}thenonlyonesolution:(a,b,c)=(1,2,1)
Commented by mathsuji last updated on 28/Apr/21
cool thanks sir
coolthankssir

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