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

$ \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$

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

### 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.**Print/PDF Version**
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