anonymous
  • anonymous
j^3 = 61 + i^6 Find all the integer solution of i and j.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
One is i=2 and j=5
anonymous
  • anonymous
What about other?
anonymous
  • anonymous
\[j^3-(i^2)^3=61\] \[(j-i^2)(j^2+ij+i^2)=61\] 61 is prime so 61(1) are the only factors hence \[j-i^2=1,j^2+ij+i^2=61\]

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anonymous
  • anonymous
Looks like 4 solution.
anonymous
  • anonymous
but we imgore the fact that it can be vise versa 1(61)
anonymous
  • anonymous
|dw:1350668697206:dw|
anonymous
  • anonymous
oh thanks i see my folly i did not factor 2
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=All+integer+solution+x^3+%3D+61+%2B+y^6
anonymous
  • anonymous
Looks like 2 and 5 only.......... But how to prove it mathematically.
anonymous
  • anonymous
\[i^4+2i^2+1+i^2(1+i^2)+i^4=61\] \[3i^4+3i^2-60=0,i^4+i^2-10=0\]
anonymous
  • anonymous
sorry \[i^4+i^2-20\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[i=\pm 2,j=(\pm2)^2+1=5\]
anonymous
  • anonymous
@Jonask you are very correct
anonymous
  • anonymous
Looks good.

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