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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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One is i=2 and j=5
What about other?
\[j^3-(i^2)^3=61\] \[(j-i^2)(j^2+ij+i^2)=61\] 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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Other answers:

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

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