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\(x^5+31=y^2\) has no integer solutions
04-02-2010, 03:27 PM
Post: #1
\(x^5+31=y^2\) has no integer solutions
We'll put more similar problems here
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04-05-2010, 12:53 PM
Post: #2
RE: \(x^5+31=y^2\) has no integer solutions
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
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