I am trying to show that the equation $x^5y + 5x^3 - xy^5 = 1$ has no solutions. Anyone has an idea on this?
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$\begingroup$ Related : math.stackexchange.com/questions/276904/… and math.stackexchange.com/questions/603104/… and math.stackexchange.com/questions/583126/… $\endgroup$ – lab bhattacharjee Jan 23 '14 at 18:43
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$$x^5y+5x^3-xy^5=y(x^5-x)+5x^3-x(y^5-y)$$
Now using Fermat's little theorem, $\displaystyle a^5-a\equiv0\pmod5$
$\displaystyle\implies$ the left hand side will always be divisible by $5$ for integer $x,y$ unlike the right one.
answered Jan 23 '14 at 17:57
