goformit100
  • goformit100
Show that there is no integer a such that a^2−3a−19 is divisible by 289.
Mathematics
chestercat
  • chestercat
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goformit100
  • goformit100
AravindG
  • AravindG
Are you sure your question right?
goformit100
  • goformit100
Ya I typed the question Now correctly, Just Accept Answer

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AravindG
  • AravindG
ok How much have you progressed in this question?
AravindG
  • AravindG
we cant JUST give YOU the answer
anonymous
  • anonymous
did you mean the factor is not integer?
goformit100
  • goformit100
Sir I am unable to Visulize this Question
goformit100
  • goformit100
goformit100
  • goformit100
NotTim
  • NotTim
ah, if only i could help.
NotTim
  • NotTim
sorry, what's an integer?
anonymous
  • anonymous
|dw:1367902685272:dw|
NotTim
  • NotTim
uh. how'd you get that?
anonymous
  • anonymous
|dw:1367902803183:dw|
goformit100
  • goformit100
Thank you Madam and Sir
anonymous
  • anonymous
then a^2-3a+(-19-289k)=0 Do you know that Ax^2+Bx+C=0? I use this B^2 -4AC A=1, B=-3, C=(-19-289k) So its be (-3)^2-4.(1).((-19-289k)
goformit100
  • goformit100
OK
anonymous
  • anonymous
9+76+1156k 85+1156k then I use 17(5+68k)
anonymous
  • anonymous
5+68k must be have 17 factor for some perfect square, 5+68k=17n 5=17n-68k 5/17=(n-4k) Found that its not integer, because there are 5/17.. even right side is integer. CMIWW @goformit100

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