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Show that there is no integer
a such that a^2−3a−19 is divisible by 289.
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goformit100
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@satellite73
AravindG
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Are you sure your question right?
goformit100
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Ya I typed the question Now correctly, Just Accept Answer
AravindG
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ok How much have you progressed in this question?
AravindG
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we cant JUST give YOU the answer
deathgrowl
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did you mean the factor is not integer?
goformit100
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Sir I am unable to Visulize this Question
goformit100
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@NotTim
goformit100
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@Machida
NotTim
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ah, if only i could help.
NotTim
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sorry, what's an integer?
Machida
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|dw:1367902685272:dw|
NotTim
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uh. how'd you get that?
Machida
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|dw:1367902803183:dw|
goformit100
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Thank you Madam and Sir
Machida
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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
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OK
Machida
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9+76+1156k
85+1156k
then I use 17(5+68k)
Machida
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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
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