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goformit100
Group Title
Show that there is no integer
a such that a^2−3a−19 is divisible by 289.
 one year ago
 one year ago
goformit100 Group Title
Show that there is no integer a such that a^2−3a−19 is divisible by 289.
 one year ago
 one year ago

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goformit100 Group TitleBest ResponseYou've already chosen the best response.0
@satellite73
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
Are you sure your question right?
 one year ago

goformit100 Group TitleBest ResponseYou've already chosen the best response.0
Ya I typed the question Now correctly, Just Accept Answer
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
ok How much have you progressed in this question?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
we cant JUST give YOU the answer
 one year ago

deathgrowl Group TitleBest ResponseYou've already chosen the best response.0
did you mean the factor is not integer?
 one year ago

goformit100 Group TitleBest ResponseYou've already chosen the best response.0
Sir I am unable to Visulize this Question
 one year ago

goformit100 Group TitleBest ResponseYou've already chosen the best response.0
@NotTim
 one year ago

goformit100 Group TitleBest ResponseYou've already chosen the best response.0
@Machida
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
ah, if only i could help.
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
sorry, what's an integer?
 one year ago

Machida Group TitleBest ResponseYou've already chosen the best response.1
dw:1367902685272:dw
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
uh. how'd you get that?
 one year ago

Machida Group TitleBest ResponseYou've already chosen the best response.1
dw:1367902803183:dw
 one year ago

goformit100 Group TitleBest ResponseYou've already chosen the best response.0
Thank you Madam and Sir
 one year ago

Machida Group TitleBest ResponseYou've already chosen the best response.1
then a^23a+(19289k)=0 Do you know that Ax^2+Bx+C=0? I use this B^2 4AC A=1, B=3, C=(19289k) So its be (3)^24.(1).((19289k)
 one year ago

Machida Group TitleBest ResponseYou've already chosen the best response.1
9+76+1156k 85+1156k then I use 17(5+68k)
 one year ago

Machida Group TitleBest ResponseYou've already chosen the best response.1
5+68k must be have 17 factor for some perfect square, 5+68k=17n 5=17n68k 5/17=(n4k) Found that its not integer, because there are 5/17.. even right side is integer. CMIWW @goformit100
 one year ago
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