To recapitulate the basics of elliptic integral theory (details here) we have
$$K = K(k) = \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}} = \int_{0}^{1}\frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$
$$E = E(k) = \int_{0}^{\pi/2}\sqrt{1 - k^{2}\sin^{2}\theta}\,d\theta = \int_{0}^{1}\frac{\sqrt{1 - k^{2}x^{2}}}{\sqrt{1 - x^{2}}}\,dx$$
We have here $ 0 \leq k \leq 1$ and the complementary modulus $ k' = \sqrt{1 - k^{2}}$ or in other words $ k^{2} + k'^{2} = 1$. Also we denote $ K' = K(k'), E' = E(k')$ and we have the Legendre's Identity $$ KE' + K'E - KK' = \frac{\pi}{2}$$
From the above arguments it follows that the function $ K'/K$ is a strictly decreasing function of $ k$. When $ k \to 0$, $ K'/K \to \infty$ and as $ k \to 1, K'/K \to 0$. Thus the function $ K'/K$ maps the interval $ (0, 1)$ to the interval $ (0, \infty)$. For any given positive number $ \alpha$ we have a unique positive $ k \in (0, 1)$ such that $ K'/K = \alpha$. Let $ p$ be any positive number and then $ p\alpha$ is also positive and therefore there is a unique number $ l \in (0, 1)$ such that $ K(l')/K(l) = p\alpha$. We traditionally denote $ L = K(l), L' = K(l')$. Therefore we can write $$\frac{L'}{L} = p\alpha = p\,\frac{K'}{K}$$
Again let's think in the following way. Suppose a number $ k \in (0, 1)$ and a positive number $ p$ is given. From $ k$ we can calculate $ K'/K$ and hence $ pK'/K$ corresponding to which there is a unique number $ l \in (0, 1)$ such that $ L'/L = pK'/K$. In this fashion $ l$ turns out to be a function of $ k$. Therefore the equation $ L'/L = pK'/K$ determines $ l$ as a function of $ k$ which is one-one and invertible. From the continuity of $ K$ the function turns out to be continuous. To derive these results formally we need the fundamental formulas: $$\frac{dk'}{dk} = -\frac{k}{k'},\,\, \frac{dK}{dk} = \frac{E - k'^{2}K}{kk'^{2}},\,\, \frac{dE}{dk} = \frac{E - K}{k}$$ (for the proofs of these results read here)
Using these we have $$\frac{dK'}{dk} = \frac{dk'}{dk}\frac{dK'}{dk'} = -\frac{k}{k'}\frac{E' - k^{2}K'}{k^{2}k'} = \frac{k^{2}K' - E'}{kk'^{2}}$$ and therefore \begin{align}\frac{d}{dk}\left(\frac{K'}{K}\right) &= \dfrac{K\dfrac{dK'}{dk} - K'\dfrac{dK}{dk}}{K^{2}} = \frac{1}{K^{2}}\left(\frac{K(k^{2}K' - E')}{kk'^{2}} - \frac{K'(E - k'^{2}K)}{kk'^{2}}\right)\notag\\ &= \frac{KK' - KE' - K'E}{kk'^{2}K^{2}} = -\frac{\pi}{2kk'^{2}K^{2}}\notag\end{align} Thus we have finally another fundamental relation $$\frac{d}{dk}\left(\frac{K'}{K}\right) = -\frac{\pi}{2kk'^{2}K^{2}}$$ which shows that $ K'/K$ is strictly decreasing. On differentiating the equation $ L'/L = pK'/K$ we get $$\frac{dl}{ll'^{2}L^{2}} = p\frac{dk}{kk'^{2}K^{2}}$$ or $$\frac{dl}{dk} = p\,\frac{ll'^{2}}{kk'^{2}}\left(\frac{L}{K}\right)^{2}$$ and therefore $ l$ is a strictly increasing function of $ k$ and maps $ (0, 1)$ to $ (0, 1)$. Also $ p <(=,>) 1 \Rightarrow l > (=,<) k$. We also see that the ratio $ L/K$ is a function of $ k, l$ and we traditionally call it the multiplier and denote by $ M_{p}(l, k)$. Thus we have $$pM_{p}^{2}(l, k) = p\left(\frac{L}{K}\right)^{2} = \frac{kk'^{2}}{ll'^{2}}\frac{dl}{dk}$$ Also it should be noted that if $ p < (=, >) 1 \Rightarrow M_{p}(l, k) > (=, <) 1$.
Next let's understand that it is necessary only to study the case when $ p$ is a positive prime. For, if for positive prime $ p$ the relation $ L'/L = pK'/K$ is leading to an algebraic relation between $ l$ and $ k$ then we can show that the same kind of relation holds when $ p$ is any positive rational number. First suppose $ p$ is composite positive integer and $ p = p_{1}p_{2}$ where both $ p_{1}, p_{2}$ are primes. Then $ L'/L = pK'/K$ can be written $ L'/L = p_{1}\Gamma'/\Gamma = p_{1}(p_{2}K'/K)$. Since the relation between $ l, \gamma$ and $ \gamma, k$ is algebraic by assumption, the relation between $ l, k$ is algebraic. The same reasoning can be extended to the case when $ p$ is a product of any number of primes. Thus the case of positive integers is handled. In case of $ p = a/b$ where $ a, b$ are positive integers we can write $ L'/L = pK'/K$ as $ bL'/L = aK'/K = \Gamma'/\Gamma$. Since the relation between $ \gamma, k$ and $ \gamma, l$ is algebraic therefore the relation between $ k, l$ is algebraic.
So from now on let's assume that $ p$ is a positive prime. The algebraic relation between $ l, k$ implied by the equation $ L'/L = pK'/K$ is called a modular equation of degree $ p$. The way Jacobi obtained these modular equations is related to the transformation of elliptic integrals. Consider the differential $$\frac{dy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}}$$ By suitable rational transformation $ y = f(x)/g(x)$ where $ f(x), g(x)$ are polynomials it is expected to reduce the differential to the form $$\frac{dx}{M\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ Thus in effect we wish to establish a relation of the form $$\frac{Mdy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ where $ y$ is a rational function of $ x$ and $ M$ is some constant. That such a transformation is possible is not so obvious. Jacobi started his Fundamenta Nova by describing this elegant theory of transformation and showed that such a transformation exists for every value of the positive prime $ p$ which is called the order of the transformation. In what follows we shall assume that $ p$ is an odd prime ($ p = 2$ is covered by the famous Landen Transformations).
The constant $ M$ is called multiplier and depends upon the values of $ l, k$ and is related to the multiplier $ M_{p}(l, k)$. To simplify the process of transformation the rational function $ y = U(x)/V(x)$ is chosen to be of very specific form. With the requirement that $ y$ vanishes with $ x$ we take $ y = xf(x)/g(x)$ where $ f(x), g(x)$ are polynomials of degrees $ (p - 1)$. Also $ f(x), g(x)$ are supposed to be even functions so that they are functions of $ x^{2}$ and then we can write $$y = \frac{xN(1, x^{2})}{D(1, x^{2})}$$ where $ N, D$ are homogeneous polynomials of degree $ (p - 1)/2$. Another condition which we impose is that the relation between $ y$ and $ x$ does not change when $ (x, y)$ is replaced by $ (1/kx, 1/ly)$. That this is possible needs to be demonstrated. First of all we can see that $$N\left(1, \frac{1}{k^{2}x^{2}}\right) = \left(\frac{1}{kx}\right)^{p - 1}N(k^{2}x^{2}, 1)$$ We can next choose $ D(1, x^{2})$ such that $ N(k^{2}x^{2}, 1) = \Delta D(1, x^{2})$ where $ \Delta$ is a constant. Thus the coefficients of $ D(1, x^{2})$ are same as those of $ N(1, x^{2})$, but in reverse order and multiplied by suitable powers of $ k$. It thus follows that $$N\left(1, \frac{1}{k^{2}x^{2}}\right) = \Delta\left(\frac{1}{kx}\right)^{p - 1}D(1, x^{2})$$ Replacing $ x$ by $ 1/kx$ we get $$N(1, x^{2}) = \Delta x^{p - 1}D\left(1, \frac{1}{k^{2}x^{2}}\right)$$ Multiplying the above two equations we get $$N(1, x^{2})N\left(1, \frac{1}{k^{2}x^{2}}\right) = \frac{\Delta^{2}}{k^{p - 1}}D(1, x^{2})D\left(1, \frac{1}{k^{2}x^{2}}\right)$$ Now let's suppose that in the defining equation if we put $ 1/kx$ in place of $ x$ then the value of $ y$ changes to $ y_{1}$. Thus $$y_{1} = \dfrac{\dfrac{1}{kx}N\left(1, \dfrac{1}{k^{2}x^{2}}\right)}{D\left(1, \dfrac{1}{k^{2}x^{2}}\right)} = \frac{1}{kx}\frac{\Delta^{2}}{k^{p - 1}}\frac{D(1, x^{2})}{N(1, x^{2})} = \frac{\Delta^{2}}{k^{p}}\frac{D(1, x^{2})}{xN(1, x^{2})} = \frac{\Delta^{2}}{k^{p}}\frac{1}{y}$$ Clearly this reduces to $ 1/ly$ provided $ l = k^{p}/\Delta^{2}$.
Hence a transformation of the form $ y = U(x)/V(x)$ with $ U$ of degree $ p$ and $ V$ of degree $ (p - 1)$ with $ U$ being odd function and $ V$ being even function is possible which remains invariant under the change of variables $ (x, y) \to (1/kx, 1/ly)$.
Next we would like to have further restrictions on the form of $ U$ and $ V$. Putting $ y = U/V$ leads to $$dy = \frac{VU' - UV'}{V^{2}}\,dx,\,\, \sqrt{(1 - y^{2})(1 - l^{2}y^{2})} = \frac{1}{V^{2}}\sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}$$ and hence $$\frac{dy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{VU' - UV'}{\sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}}\,dx$$ If it is possible that $ V + U = (1 + x)A^{2}, V - U = (1 - x)B^{2}$ for some polynomials $ A, B$ then we can see that $$1 + y = (1 + x)A^{2}/V,\, 1 - y = (1 - x)B^{2}/V$$ and by invariance of relation between $ x$ and $ y$ under the transformation $$ (x, y) \to (1/kx, 1/ky)$$ we can see that we must also have the relations $$1 + ly = (1 + kx)C^{2}/V,\, 1 - ly = (1 - kx)D^{2}/V$$ or $$V +lU = (1 + kx)C^{2},\, V - lU = (1 - kx)D^{2}$$ Now we can see that if $ (x - \alpha)$ is a factor of $ A$ then $ (x - \alpha)^{2}$ divides $ V + U$ and hence $ (x - \alpha)$ divides $ (V + U)U' - U(V + U)' = VU' - UV'$. From the same logic it follows that $ A, B, C, D$ each divide $ VU' - UV'$ and since the degrees of $ VU' - UV'$ and $ ABCD$ are same (equal to $ (2p - 2)$) it follows that the ratio $ ABCD/(VU' - UV')$ is a constant which we denote by $ M$. Thus we arrive at $$\frac{Mdy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ The idea is now to choose polynomials $ A, B$ in a suitable form. We keep $ A = P + Qx, B = P - Qx$ such that $ P, Q$ are even functions of $ x$ and the degree of $ P \pm Qx$ is $ (p - 1)/2$. In general if $ p = (4n - 1)$ then the degrees of $ P, Q$ are $ (2n - 2)$ and if $ p = 4n + 1$, degree of $ P$ is $ 2n$ and that of $ Q$ is $ (2n - 2)$.
Thus we arrive at the following transformation $$\frac{1 - y}{1 + y} = \frac{1 - x}{1 + x}\frac{(P - Qx)^{2}}{(P + Qx)^{2}}$$ so that $$y = \frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}$$ and then we obtain \begin{align}1 - y &= \frac{(1 - x)(P - Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\\ 1 + y &= \frac{(1 + x)(P + Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\\ 1 - ly &= \frac{(1 - kx)(P' - Q'x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\\ 1 + ly &= \frac{(1 + kx)(P' + Q'x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\end{align} The actual methodology of finding the polynomials $ U, V$ (or $ P, Q$) would be dealt with in next post by showing examples with $ p = 3$ and $ p = 5$. From the theoretical investigation above it is clear that the process of finding these polynomials (i.e. their coefficients) is totally algebraical and hence the relation between $ k$ and $ l$ is algebraic (note that $ l = k^{p}/\Delta^{2}$, and that $ \Delta$ itself would be an algebraic function of $ l, k$). It should also be observed that the multiplier $ M$ is also an algebraical function of $ l$ and $ k$ (calculation of $ M$ is easy if we observe that $ 1/M = dy/dx \text{ at } x = 0$).
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We have here $ 0 \leq k \leq 1$ and the complementary modulus $ k' = \sqrt{1 - k^{2}}$ or in other words $ k^{2} + k'^{2} = 1$. Also we denote $ K' = K(k'), E' = E(k')$ and we have the Legendre's Identity $$ KE' + K'E - KK' = \frac{\pi}{2}$$
The Function $ K'/K$
From the definition it is clear that $ K$ is a strictly increasing function of $ k$ and moreover $ K \to \infty$ as $ k \to 1$. Thus function $ K(k)$ maps interval $ [0, 1)$ to $ [\pi/2, \infty)$. Similarly $ E$ is strictly decreasing and maps $ [0, 1]$ to $ [1, \pi/2]$. Therefore it follows that $ K'$ is a strictly decreasing function of $ k$ and $ E'$ is a strictly increasing function of $ k$.From the above arguments it follows that the function $ K'/K$ is a strictly decreasing function of $ k$. When $ k \to 0$, $ K'/K \to \infty$ and as $ k \to 1, K'/K \to 0$. Thus the function $ K'/K$ maps the interval $ (0, 1)$ to the interval $ (0, \infty)$. For any given positive number $ \alpha$ we have a unique positive $ k \in (0, 1)$ such that $ K'/K = \alpha$. Let $ p$ be any positive number and then $ p\alpha$ is also positive and therefore there is a unique number $ l \in (0, 1)$ such that $ K(l')/K(l) = p\alpha$. We traditionally denote $ L = K(l), L' = K(l')$. Therefore we can write $$\frac{L'}{L} = p\alpha = p\,\frac{K'}{K}$$
Again let's think in the following way. Suppose a number $ k \in (0, 1)$ and a positive number $ p$ is given. From $ k$ we can calculate $ K'/K$ and hence $ pK'/K$ corresponding to which there is a unique number $ l \in (0, 1)$ such that $ L'/L = pK'/K$. In this fashion $ l$ turns out to be a function of $ k$. Therefore the equation $ L'/L = pK'/K$ determines $ l$ as a function of $ k$ which is one-one and invertible. From the continuity of $ K$ the function turns out to be continuous. To derive these results formally we need the fundamental formulas: $$\frac{dk'}{dk} = -\frac{k}{k'},\,\, \frac{dK}{dk} = \frac{E - k'^{2}K}{kk'^{2}},\,\, \frac{dE}{dk} = \frac{E - K}{k}$$ (for the proofs of these results read here)
Using these we have $$\frac{dK'}{dk} = \frac{dk'}{dk}\frac{dK'}{dk'} = -\frac{k}{k'}\frac{E' - k^{2}K'}{k^{2}k'} = \frac{k^{2}K' - E'}{kk'^{2}}$$ and therefore \begin{align}\frac{d}{dk}\left(\frac{K'}{K}\right) &= \dfrac{K\dfrac{dK'}{dk} - K'\dfrac{dK}{dk}}{K^{2}} = \frac{1}{K^{2}}\left(\frac{K(k^{2}K' - E')}{kk'^{2}} - \frac{K'(E - k'^{2}K)}{kk'^{2}}\right)\notag\\ &= \frac{KK' - KE' - K'E}{kk'^{2}K^{2}} = -\frac{\pi}{2kk'^{2}K^{2}}\notag\end{align} Thus we have finally another fundamental relation $$\frac{d}{dk}\left(\frac{K'}{K}\right) = -\frac{\pi}{2kk'^{2}K^{2}}$$ which shows that $ K'/K$ is strictly decreasing. On differentiating the equation $ L'/L = pK'/K$ we get $$\frac{dl}{ll'^{2}L^{2}} = p\frac{dk}{kk'^{2}K^{2}}$$ or $$\frac{dl}{dk} = p\,\frac{ll'^{2}}{kk'^{2}}\left(\frac{L}{K}\right)^{2}$$ and therefore $ l$ is a strictly increasing function of $ k$ and maps $ (0, 1)$ to $ (0, 1)$. Also $ p <(=,>) 1 \Rightarrow l > (=,<) k$. We also see that the ratio $ L/K$ is a function of $ k, l$ and we traditionally call it the multiplier and denote by $ M_{p}(l, k)$. Thus we have $$pM_{p}^{2}(l, k) = p\left(\frac{L}{K}\right)^{2} = \frac{kk'^{2}}{ll'^{2}}\frac{dl}{dk}$$ Also it should be noted that if $ p < (=, >) 1 \Rightarrow M_{p}(l, k) > (=, <) 1$.
Jacobi's Transformation Theory
Jacobi showed that the relation between $ l$ and $ k$ is algebraic when $ p$ is rational. First of all as an example we can take $ p = 2$. Then by Landen's transformation $ l = (1 - k')/(1 + k') < k$ we see that $ L'/L = 2K'/K$. Clearly here the relation between $ l$ and $ k$ is algebraic and is also given by $ k = 2\sqrt{l}/(1 + l)$. Also $$\frac{dl}{dk} = \frac{(1 + k')(k/k') + (1 - k')(k/k')}{(1 + k')^{2}} = \frac{2k}{k'(1 + k')^{2}}$$ and therefore $$2M_{2}^{2}(l, k) = \frac{kk'^{2}}{ll'^{2}}\frac{dl}{dk} = kk'^{2}\frac{1 + k'}{1 - k'}\frac{(1 + k')^{2}}{4k'}\frac{2k}{k'(1 + k')^{2}} = \frac{k^{2}}{2}\frac{(1 + k')^{2}}{1 - k'^{2}}$$ so that $ L/K = M_{2}(l, k) = (1 + k')/2 = 1/(1 + l)$ as should have been the case.Next let's understand that it is necessary only to study the case when $ p$ is a positive prime. For, if for positive prime $ p$ the relation $ L'/L = pK'/K$ is leading to an algebraic relation between $ l$ and $ k$ then we can show that the same kind of relation holds when $ p$ is any positive rational number. First suppose $ p$ is composite positive integer and $ p = p_{1}p_{2}$ where both $ p_{1}, p_{2}$ are primes. Then $ L'/L = pK'/K$ can be written $ L'/L = p_{1}\Gamma'/\Gamma = p_{1}(p_{2}K'/K)$. Since the relation between $ l, \gamma$ and $ \gamma, k$ is algebraic by assumption, the relation between $ l, k$ is algebraic. The same reasoning can be extended to the case when $ p$ is a product of any number of primes. Thus the case of positive integers is handled. In case of $ p = a/b$ where $ a, b$ are positive integers we can write $ L'/L = pK'/K$ as $ bL'/L = aK'/K = \Gamma'/\Gamma$. Since the relation between $ \gamma, k$ and $ \gamma, l$ is algebraic therefore the relation between $ k, l$ is algebraic.
So from now on let's assume that $ p$ is a positive prime. The algebraic relation between $ l, k$ implied by the equation $ L'/L = pK'/K$ is called a modular equation of degree $ p$. The way Jacobi obtained these modular equations is related to the transformation of elliptic integrals. Consider the differential $$\frac{dy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}}$$ By suitable rational transformation $ y = f(x)/g(x)$ where $ f(x), g(x)$ are polynomials it is expected to reduce the differential to the form $$\frac{dx}{M\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ Thus in effect we wish to establish a relation of the form $$\frac{Mdy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ where $ y$ is a rational function of $ x$ and $ M$ is some constant. That such a transformation is possible is not so obvious. Jacobi started his Fundamenta Nova by describing this elegant theory of transformation and showed that such a transformation exists for every value of the positive prime $ p$ which is called the order of the transformation. In what follows we shall assume that $ p$ is an odd prime ($ p = 2$ is covered by the famous Landen Transformations).
The constant $ M$ is called multiplier and depends upon the values of $ l, k$ and is related to the multiplier $ M_{p}(l, k)$. To simplify the process of transformation the rational function $ y = U(x)/V(x)$ is chosen to be of very specific form. With the requirement that $ y$ vanishes with $ x$ we take $ y = xf(x)/g(x)$ where $ f(x), g(x)$ are polynomials of degrees $ (p - 1)$. Also $ f(x), g(x)$ are supposed to be even functions so that they are functions of $ x^{2}$ and then we can write $$y = \frac{xN(1, x^{2})}{D(1, x^{2})}$$ where $ N, D$ are homogeneous polynomials of degree $ (p - 1)/2$. Another condition which we impose is that the relation between $ y$ and $ x$ does not change when $ (x, y)$ is replaced by $ (1/kx, 1/ly)$. That this is possible needs to be demonstrated. First of all we can see that $$N\left(1, \frac{1}{k^{2}x^{2}}\right) = \left(\frac{1}{kx}\right)^{p - 1}N(k^{2}x^{2}, 1)$$ We can next choose $ D(1, x^{2})$ such that $ N(k^{2}x^{2}, 1) = \Delta D(1, x^{2})$ where $ \Delta$ is a constant. Thus the coefficients of $ D(1, x^{2})$ are same as those of $ N(1, x^{2})$, but in reverse order and multiplied by suitable powers of $ k$. It thus follows that $$N\left(1, \frac{1}{k^{2}x^{2}}\right) = \Delta\left(\frac{1}{kx}\right)^{p - 1}D(1, x^{2})$$ Replacing $ x$ by $ 1/kx$ we get $$N(1, x^{2}) = \Delta x^{p - 1}D\left(1, \frac{1}{k^{2}x^{2}}\right)$$ Multiplying the above two equations we get $$N(1, x^{2})N\left(1, \frac{1}{k^{2}x^{2}}\right) = \frac{\Delta^{2}}{k^{p - 1}}D(1, x^{2})D\left(1, \frac{1}{k^{2}x^{2}}\right)$$ Now let's suppose that in the defining equation if we put $ 1/kx$ in place of $ x$ then the value of $ y$ changes to $ y_{1}$. Thus $$y_{1} = \dfrac{\dfrac{1}{kx}N\left(1, \dfrac{1}{k^{2}x^{2}}\right)}{D\left(1, \dfrac{1}{k^{2}x^{2}}\right)} = \frac{1}{kx}\frac{\Delta^{2}}{k^{p - 1}}\frac{D(1, x^{2})}{N(1, x^{2})} = \frac{\Delta^{2}}{k^{p}}\frac{D(1, x^{2})}{xN(1, x^{2})} = \frac{\Delta^{2}}{k^{p}}\frac{1}{y}$$ Clearly this reduces to $ 1/ly$ provided $ l = k^{p}/\Delta^{2}$.
Hence a transformation of the form $ y = U(x)/V(x)$ with $ U$ of degree $ p$ and $ V$ of degree $ (p - 1)$ with $ U$ being odd function and $ V$ being even function is possible which remains invariant under the change of variables $ (x, y) \to (1/kx, 1/ly)$.
Next we would like to have further restrictions on the form of $ U$ and $ V$. Putting $ y = U/V$ leads to $$dy = \frac{VU' - UV'}{V^{2}}\,dx,\,\, \sqrt{(1 - y^{2})(1 - l^{2}y^{2})} = \frac{1}{V^{2}}\sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}$$ and hence $$\frac{dy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{VU' - UV'}{\sqrt{(V^{2} - U^{2})(V^{2} - l^{2}U^{2})}}\,dx$$ If it is possible that $ V + U = (1 + x)A^{2}, V - U = (1 - x)B^{2}$ for some polynomials $ A, B$ then we can see that $$1 + y = (1 + x)A^{2}/V,\, 1 - y = (1 - x)B^{2}/V$$ and by invariance of relation between $ x$ and $ y$ under the transformation $$ (x, y) \to (1/kx, 1/ky)$$ we can see that we must also have the relations $$1 + ly = (1 + kx)C^{2}/V,\, 1 - ly = (1 - kx)D^{2}/V$$ or $$V +lU = (1 + kx)C^{2},\, V - lU = (1 - kx)D^{2}$$ Now we can see that if $ (x - \alpha)$ is a factor of $ A$ then $ (x - \alpha)^{2}$ divides $ V + U$ and hence $ (x - \alpha)$ divides $ (V + U)U' - U(V + U)' = VU' - UV'$. From the same logic it follows that $ A, B, C, D$ each divide $ VU' - UV'$ and since the degrees of $ VU' - UV'$ and $ ABCD$ are same (equal to $ (2p - 2)$) it follows that the ratio $ ABCD/(VU' - UV')$ is a constant which we denote by $ M$. Thus we arrive at $$\frac{Mdy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ The idea is now to choose polynomials $ A, B$ in a suitable form. We keep $ A = P + Qx, B = P - Qx$ such that $ P, Q$ are even functions of $ x$ and the degree of $ P \pm Qx$ is $ (p - 1)/2$. In general if $ p = (4n - 1)$ then the degrees of $ P, Q$ are $ (2n - 2)$ and if $ p = 4n + 1$, degree of $ P$ is $ 2n$ and that of $ Q$ is $ (2n - 2)$.
Thus we arrive at the following transformation $$\frac{1 - y}{1 + y} = \frac{1 - x}{1 + x}\frac{(P - Qx)^{2}}{(P + Qx)^{2}}$$ so that $$y = \frac{x(P^{2} + 2PQ + Q^{2}x^{2})}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}$$ and then we obtain \begin{align}1 - y &= \frac{(1 - x)(P - Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\\ 1 + y &= \frac{(1 + x)(P + Qx)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\\ 1 - ly &= \frac{(1 - kx)(P' - Q'x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\\ 1 + ly &= \frac{(1 + kx)(P' + Q'x)^{2}}{P^{2} + 2PQx^{2} + Q^{2}x^{2}}\notag\end{align} The actual methodology of finding the polynomials $ U, V$ (or $ P, Q$) would be dealt with in next post by showing examples with $ p = 3$ and $ p = 5$. From the theoretical investigation above it is clear that the process of finding these polynomials (i.e. their coefficients) is totally algebraical and hence the relation between $ k$ and $ l$ is algebraic (note that $ l = k^{p}/\Delta^{2}$, and that $ \Delta$ itself would be an algebraic function of $ l, k$). It should also be observed that the multiplier $ M$ is also an algebraical function of $ l$ and $ k$ (calculation of $ M$ is easy if we observe that $ 1/M = dy/dx \text{ at } x = 0$).
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y=x*f(x)/g(x) Why f(x),g(x) should be polynomials of degrees p-1?
pokhkitro
December 20, 2021 at 6:09 PM