\(x^5+31=y^2\) has no integer solutions

[elim]

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[elim]

Sol. Since \((2m)^5+31 \equiv 3 \pmod{4}\), \(x\) cannot be even
For any odd integer \(x\), we have \(x^5+31\pmod {11} \in \left\{8,10\right \} \) but \(y^2\pmod{11} \in \left\{0,1,3,4,5,9\right \}\)
So \(x^5+31=y^2\) has no integer solutions