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

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

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

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

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

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

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

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

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

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

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

Machida
 one year ago
Best ResponseYou've already chosen the best response.1then 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)

Machida
 one year ago
Best ResponseYou've already chosen the best response.19+76+1156k 85+1156k then I use 17(5+68k)

Machida
 one year ago
Best ResponseYou've already chosen the best response.15+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
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