I'd appreciate some help (a hint) for the following. If $x,y>1$ are so that $2x^2-1=y^{15}$ then $x$ is a multiple of $5$.
Don't know if this helps but the equation can be rewritten as $2x^2=(y^5+1)(y^2-y+1)(y^8+y^7-y^5-y^4-y^3+y+1)$
2026-04-24 01:14:26.1776993266
15th power Diophantine equation
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