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I have formula for computing the j-invariant but I was wondering if given number $j$, is there a formula for getting a curve $y^2=x^3+a_2x^2+a_4x+a_6$ with j-invariant j?

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I found it!

\begin{eqnarray*} Y^2 + XY = X^3 -\dfrac{36}{j-1728}X-\dfrac{1}{j-1728} \end{eqnarray*}

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  • $\begingroup$ This construction works unless $j=0$ or $j=1728$. But $y^2+y=x^3$ has $j=0$, and $y^2=x^3+x$ has $j=1728$. $\endgroup$ – Álvaro Lozano-Robledo Apr 15 '13 at 13:35

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