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sauravshakyaBest ResponseYou've already chosen the best response.1
One is i=2 and j=5
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
What about other?
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
\[j^3(i^2)^3=61\] \[(ji^2)(j^2+ij+i^2)=61\] 61 is prime so 61(1) are the only factors hence \[ji^2=1,j^2+ij+i^2=61\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Looks like 4 solution.
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
but we imgore the fact that it can be vise versa 1(61)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
dw:1350668697206:dw
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
oh thanks i see my folly i did not factor 2
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=All+integer+solution+x^3+%3D+61+%2B+y^6
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Looks like 2 and 5 only.......... But how to prove it mathematically.
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
\[i^4+2i^2+1+i^2(1+i^2)+i^4=61\] \[3i^4+3i^260=0,i^4+i^210=0\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.2
\[i=\pm 2,j=(\pm2)^2+1=5\]
 one year ago

ZekariasBest ResponseYou've already chosen the best response.0
@Jonask you are very correct
 one year ago
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