$$x^5+31=y^2$$ has no integer solutions - Printable Version +- OMath! (http://math.elinkage.net) +-- Forum: Math Forums (/forumdisplay.php?fid=4) +--- Forum: Number Theory (/forumdisplay.php?fid=7) +--- Thread: $$x^5+31=y^2$$ has no integer solutions (/showthread.php?tid=37) $$x^5+31=y^2$$ has no integer solutions - elim - 04-02-2010 03:27 PM We'll put more similar problems here RE: $$x^5+31=y^2$$ has no integer solutions - elim - 04-05-2010 12:53 PM 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