<s>[INST] Diophantine equation: $y^2=1+12x+16x^2$ The diophantine equation $$y^2=1+12x+16x^2$$ only has solutions $x=0, y=\pm1$ according to wolfram alpha. How would I go about proving these are the only solutions?
Similarly the equation $$y^2=5+12x+16x^2$$ has solutions $x=-1, y=\pm3$.
Is there a general method with regards to these types of equations? Thanks.
 [/INST] $1+12x+16x^2$ is closest to $(4x+1)^2$ and $(4x+2)^2$ regardless of $x\geq0$ or $x<0$, as it is easy to see the distance from $(4x)^2$ and $(4x+3)^2$ to $1+12x+16x^2$ are strictly larger. You need to check both equality and finds out it has integer solution only when $x=0$
Similarly,
$5+12x+16x^2$ is closest to $(4x+1)^2$ and $(4x+2)^2$ as well, the equality holds only when $5+12x+16x^2=(4x+1)^2$ and $x=-1$.
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