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

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did you mean the factor is not integer?

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1367902685272:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1367902803183:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.09+76+1156k 85+1156k then I use 17(5+68k)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.05+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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