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j^3 = 61 + i^6 Find all the integer solution of i and j.
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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 \[j-i^2=1,j^2+ij+i^2=61\]
Looks like 4 solution.
but we imgore the fact that it can be vise versa 1(61)
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