So the equation I am trying to solve is $x^2=y^4-77$
So far I have rearranged and factorised the equation to get: $$(y^2-x)(y^2+x)=77$$
But I am really unsure of how to solve it from here. Thanks in advance
2026-04-01 17:08:44.1775063324
Solving diophantine equations
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$(y^2−x)(y^2+x)=77$
$77 = 7*11$ so $(y^2+x=11)$ and $(y^2-x=7)$
$(y^2+x)-(y^2−x)=2x$ = $11-7 = 4$
Hence $x=2$ and $y^2 + 2 = 11$ so $y=3$
$x$ and $y$ can take negative values as well so range of solutions is $x=\pm 2$ , $y=\pm 3$