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goformit100

  • 2 years ago

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

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  1. goformit100
    • 2 years ago
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    @satellite73

  2. AravindG
    • 2 years ago
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    Are you sure your question right?

  3. goformit100
    • 2 years ago
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    Ya I typed the question Now correctly, Just Accept Answer

  4. AravindG
    • 2 years ago
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    ok How much have you progressed in this question?

  5. AravindG
    • 2 years ago
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    we cant JUST give YOU the answer

  6. deathgrowl
    • 2 years ago
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    did you mean the factor is not integer?

  7. goformit100
    • 2 years ago
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    Sir I am unable to Visulize this Question

  8. goformit100
    • 2 years ago
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    @NotTim

  9. goformit100
    • 2 years ago
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    @Machida

  10. NotTim
    • 2 years ago
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    ah, if only i could help.

  11. NotTim
    • 2 years ago
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    sorry, what's an integer?

  12. Machida
    • 2 years ago
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    |dw:1367902685272:dw|

  13. NotTim
    • 2 years ago
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    uh. how'd you get that?

  14. Machida
    • 2 years ago
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    |dw:1367902803183:dw|

  15. goformit100
    • 2 years ago
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    Thank you Madam and Sir

  16. Machida
    • 2 years ago
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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)

  17. goformit100
    • 2 years ago
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    OK

  18. Machida
    • 2 years ago
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    9+76+1156k 85+1156k then I use 17(5+68k)

  19. Machida
    • 2 years ago
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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 @goformit100

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