If we consider mod 7, then $2\equiv y^3$ (mod 7). How do I conclude that this is not possible?
Prove there are no integers $x,y$ satisfying $7x^{10}+2=y^3$
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number-theory
modular-arithmetic
1 Answers
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If you explicitly cube all seven possible values $\pmod 7$ you get $\{0,1,1,6,1,6,6\}$ (actually, you need only cube $2$ and $3$ because obviously neither $\pm 1^3$ nor $0^3$ will equal $2$, and $(-a)^3 = -a^3 \pmod 7$.
By inspection, non of those values is $2$.