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

Mathematics
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Other answers:

ok How much have you progressed in this question?
we cant JUST give YOU the answer
did you mean the factor is not integer?
Sir I am unable to Visulize this Question
ah, if only i could help.
sorry, what's an integer?
|dw:1367902685272:dw|
uh. how'd you get that?
|dw:1367902803183:dw|
Thank you Madam and Sir
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)
OK
9+76+1156k 85+1156k then I use 17(5+68k)
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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