# Elementary Approach to Modular Equations: Jacobi's Transformation Theory 2

In this post we will apply the technique described in previous post to obtain modular equations of degree $3$ and $5$.

### Cubic Transformation $p = 3$

Since we have $p = 4\cdot 1 - 1$, the degrees of $P$ and $Q$ are both $2\cdot 1 - 2 = 0$ and hence they both are constants. Since
$\displaystyle y = \frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}$
it is clear that the function $y$ depends only on ratio $Q/P$ and hence we can take $P = 1$ and $Q = \alpha$. Thus we have
$\displaystyle y = \frac{x(1 + 2\alpha + \alpha^{2}x^{2})}{1 + 2\alpha x^{2} + \alpha^{2}x^{2}} = x\cdot \frac{(1 + 2\alpha) + \alpha^{2}x^{2}}{1 + \alpha(\alpha + 2)x^{2}}$

The requirement of invariance under the transformation $(x, y) \to (1/kx, 1/ly)$ leads to the following conditions
$\displaystyle l = \frac{k^{3}}{\Delta^{2}}$
and
$\displaystyle (1 + 2\alpha)k^{2}x^{2} + \alpha^{2} = \Delta\{1 + \alpha(\alpha + 2)x^{2}\}$

Thus we have the following equations
$\displaystyle l = \frac{k^{3}}{\Delta^{2}}$
$\alpha^{2} = \Delta$
$k^{2}(2\alpha + 1) = \Delta \alpha(\alpha + 2)$
and since $1/M = dy/dx$ at $x = 0$ we have $M = 1/(2\alpha + 1)$. Clearly we have then
$\displaystyle k^{2} = \frac{\alpha^{3}(2 + \alpha)}{2\alpha + 1},\,\, l^{2} = \alpha \left(\frac{2 + \alpha}{2\alpha + 1}\right)^{3}$

Eliminating $\alpha$ from the above equation is bit tricky. After some manipulations we have
$\displaystyle k'^{2} = \frac{(1 - \alpha)(1 + \alpha)^{3}}{2\alpha + 1},\,\, l'^{2} = \frac{(1 + \alpha)(1 - \alpha)^{3}}{(2\alpha + 1)^{3}}$

Clearly the pattern is now obvious if we multiply the expressions for $k, l$ and (similarly for $k', l'$)
$\displaystyle \sqrt{kl} = \frac{\alpha(2 + \alpha)}{2\alpha + 1},\,\, \sqrt{k'l'} = \frac{1 - \alpha^{2}}{2\alpha + 1}$
and thus we obtain the modular equation of degree $3$:
$\sqrt{kl} + \sqrt{k'l'} = 1$.

Using the variable $u = k^{1/4}, v = l^{1/4}$ we can obtain an expression devoid of algebraic irrationalities. Clearly we have $\alpha^{4} = k^{3}/l$ so that $\alpha = u^{3}/v$ and since $\sqrt{kl} = \alpha(2 + \alpha)/(2\alpha + 1)$ it follows that $u^{2}v^{2} = \alpha(2 + \alpha)/(2\alpha + 1)$. Putting the value of $\alpha$ in this equation we get
$\displaystyle u^{2}v^{2} = \dfrac{\dfrac{u^{3}}{v}\left(2 + \dfrac{u^{3}}{v}\right)}{\dfrac{2u^{3}}{v} + 1} = \frac{u^{3}(2v + u^{3})}{v(2u^{3} + v)}$
$\Rightarrow u(u^{3} + 2v) = v^{3}(2u^{3} + v)$
and thus finally
$u^{4} - v^{4} + 2uv(1 - u^{2}v^{2}) = 0$

### Quintic Transformation $p = 5$

Here we have $p = 5 = 4\cdot 1 + 1$, hence the degree of $P$ is $1$ and that of $Q$ is zero. Thus we can take $P = 1 + \beta x^{2}$ and $Q = \alpha$ (note that the degrees being talked of are degrees when $P, Q$ are expressed as polynomials in $x^{2}$).

We have now
$\displaystyle y = \frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}$
so that in this case
$\displaystyle y = \frac{x\{(2\alpha + 1) + (2\alpha\beta + 2\beta + \alpha^{2})x^{2} + \beta^{2}x^{4}\}}{1 + (2\beta + 2\alpha + \alpha^{2})x^{2} + (\beta^{2} + 2\alpha\beta)x^{4}}$

This relation is invariant under the transformation $(x, y) \to (1/kx, 1/ly)$ so we have
$\displaystyle l = \frac{k^{5}}{\Delta^{2}}\,\,\,\cdots (1)$
$\beta^{2} = \Delta\,\,\,\cdots (2)$
$k^{2}(2\alpha\beta + 2\beta + \alpha^{2}) = \Delta(2\beta + 2\alpha + \alpha^{2})\,\,\,\cdots (3)$
$k^{4}(2\alpha + 1) = \Delta(\beta^{2} + 2\alpha\beta)\,\,\,\cdots (4)$

Like in the case of cubic transformation we have $M = 1/(2\alpha + 1)$.

Obtaining a relation between $k, l$ by eliminating $\alpha, \beta, \Delta$ is reasonably cumbersome and it becomes manageable when we use the associated variables $u = k^{1/4}, v = l^{1/4}$. Using these variables and equations (1) and (2) we have $\beta = \sqrt{\Delta} = u^{5}/v$ and putting these values in equation (4) we get
$\displaystyle u^{16}(2\alpha + 1) = \frac{u^{10}}{v^{2}}\left(\frac{u^{10}}{v^{2}} + 2\alpha\cdot \frac{u^{5}}{v}\right)$
or
$(2\alpha + 1)uv^{4} = u^{5} + 2\alpha v \Rightarrow 2\alpha v(1 - uv^{3}) = u(v^{4} - u^{4})$
$\displaystyle \Rightarrow 2\alpha = \frac{u(v^{4} - u^{4})}{v(1 - uv^{3})}\,\,\,\cdots (5)$

Again using equation (3) we get
$\displaystyle (v^{2} - u^{2})(2\beta + \alpha^{2}) = u^{2}(1 - u^{3}v)(2\alpha) = \frac{u^{3}}{v}\frac{(v^{4} - u^{4})(1 - u^{3}v)}{1 - uv^{3}}$
$\displaystyle \Rightarrow 2\beta + \alpha^{2} = \frac{u^{3}}{v}\frac{(v^{2} + u^{2})(1 - u^{3}v)}{1 - uv^{3}}$
$\displaystyle \Rightarrow \alpha^{2} = \frac{u^{3}}{v}\left\{\frac{(v^{2} + u^{2})(1 - u^{3}v)}{1 - uv^{3}} - 2u^{2}\right\}$
$\displaystyle \Rightarrow \alpha^{2} = \frac{u^{3}}{v}\frac{(v^{2} - u^{2})(1 + u^{3}v)}{1 - uv^{3}}\,\,\,\cdots (6)$

Dividing (6) by (5) we get
$\displaystyle 2\alpha = \frac{4u^{2}(1 + u^{3}v)}{v^{2} + u^{2}}$
Comparing this with (5) we get
$4uv(1 - uv^{3})(1 + u^{3}v) - (v^{2} + u^{2})(v^{4} - u^{4}) = 0$
and on simplifying further we have
$u^{6} - v^{6} + 5u^{2}v^{2}(u^{2} - v^{2}) + 4uv(1 - u^{4}v^{4}) = 0$

The reader must have understood by now that this technique of finding modular equations is quite unsuitable for larger values of $p$. In fact Jacobi treated only the cubic and quintic transformations in his Fundamenta Nova and Arthur Cayley extended this approach to $p = 7$ in his An Elementary Treatise on Elliptic Functions. Our exposition here is based on Cayley's book. For higher values of $p$ Jacobi provided the transformation in a form which contains elliptic functions of $K/p$ and thereby obtained equations $L = K/(pM), L' = K'/M$ from which follows the relation $L'/L = pK'/K$. This we study in the next post.