I have to prove that the equation $$15x^3+13y^3=101$$ does not have any solutions over the integers. Can you provide any hint?
2026-04-04 06:10:14.1775283014
A diophantine equation with cubes that doesn't have solutions
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$\pmod{7}$, a perfect cube $\in\{0,1,6\}$. That implies: $$ 15x^3+13y^3 \equiv x^3+(-y)^3 \in \{0,1,2,5,6\} \pmod{7}$$ while $101\equiv\color{red}{3}\pmod{7}$.