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sauravshakya

  • 2 years ago

j^3 = 61 + i^6 Find all the integer solution of i and j.

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  1. sauravshakya
    • 2 years ago
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    One is i=2 and j=5

  2. sauravshakya
    • 2 years ago
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    What about other?

  3. Jonask
    • 2 years ago
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    \[j^3-(i^2)^3=61\] \[(j-i^2)(j^2+ij+i^2)=61\] 61 is prime so 61(1) are the only factors hence \[j-i^2=1,j^2+ij+i^2=61\]

  4. sauravshakya
    • 2 years ago
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    Looks like 4 solution.

  5. Jonask
    • 2 years ago
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    but we imgore the fact that it can be vise versa 1(61)

  6. sauravshakya
    • 2 years ago
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    |dw:1350668697206:dw|

  7. Jonask
    • 2 years ago
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    oh thanks i see my folly i did not factor 2

  8. sauravshakya
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=All+integer+solution+x^3+%3D+61+%2B+y^6

  9. sauravshakya
    • 2 years ago
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    Looks like 2 and 5 only.......... But how to prove it mathematically.

  10. Jonask
    • 2 years ago
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    \[i^4+2i^2+1+i^2(1+i^2)+i^4=61\] \[3i^4+3i^2-60=0,i^4+i^2-10=0\]

  11. Jonask
    • 2 years ago
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    sorry \[i^4+i^2-20\]

  12. Jonask
    • 2 years ago
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    yes

  13. Zekarias
    • 2 years ago
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    ok

  14. Jonask
    • 2 years ago
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    \[i=\pm 2,j=(\pm2)^2+1=5\]

  15. Zekarias
    • 2 years ago
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    @Jonask you are very correct

  16. sauravshakya
    • 2 years ago
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    Looks good.

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