Is the following true?
"$$\frac{\pi}{4}=\arctan \frac{1}{2}+\arctan \frac{1}{3}$$$$\frac{\pi}{4}=2\arctan \frac{1}{2}-\arctan \frac{1}{7}$$$$\frac{\pi}{4}=2\arctan \frac{1}{3}+\arctan \frac{1}{7}$$$$\frac{\pi}{4}=4\arctan \frac{1}{5}-\arctan \frac{1}{239}$$
are only solutions of $$\frac{\pi}{4}=k\arctan \frac{1}{m}+l\arctan \frac{1}{n}\tag{$\star$}$$ where $k,l,m,n$ be integers such that $kl\not=0, 0\lt m\lt n$."
Motivation : I found the following question in a book. :
Find every rational number $x$ such that $$\frac{\pi}{4}=\arctan\frac{m}{n}+\arctan{x}$$ where $m,n$ are integers.
By using addition theorem of tangent, we can get $x=\frac{n-m}{n+m}.$
This got me interested in $(\star)$, but I can neither find the other solutions nor prove that there is no other solution. Can anyone help?
I've just been able to prove that there are no other solutions except these four solutions.
Note that in the following I change the condition to the following : $$\frac{\pi}{4}=k\arctan\frac{1}{m}+l\arctan\frac{1}{n}\ \ \ \ \cdots(\star)$$ where $k\gt0, l\gt0, 0\lt|m|\lt|n|$.
First, let's prove that $k$ and $l$ are coprime with each other.
If $(k,l)=d\gt 1$, then dividing the both sides of $(\star)$ by $d$, we get $$\frac{\pi}{4d}=\frac kd\arctan\frac{1}{m}+\frac ld\arctan\frac1n.$$ By taking $\tan$ to the both sides, we know that the left side is an irrational number and the right side a rational number, which is a contradiction. (Note that Ivan Niven proved that $\tan\frac{\pi}{4d}\ (d\gt 1)$ is an irrational number.)
From $(\star)$, since the real part of $(m+i)^k(n+1)^l$ is equal to its imaginary part, we know that $R=(1-i)(m+i)^k(n+i)^l$ is a real number. Hence, we get $$(1-i)(m+i)^k(n+i)^l=(1+i)(m-i)^k(n-i)^k\ \ \ \ \cdots(\star\star).$$ Here, let us define $\mu= \begin{cases} 0, & \text{if $m$ is even} \\ 1, & \text{if $m$ is odd} \\ \end{cases},$$\ \nu= \begin{cases} 0, & \text{if $n$ is even} \\ 1, & \text{if $n$ is odd} \\ \end{cases}$.
Letting $m+i=(1+i)^{\mu}(a+bi)$, then taking the norm of the both sides gives $m^2+1=2^{\mu}(a^2+b^2)$. Hence, by the definition of $\mu$, we know that $a,b$ are integers and that $a^2+b^2$ is odd. So, $a+bi$ is not divided by $1+i$. By the same argument as above, letting $n+i=(1+i)^{\nu}(c+di)$ tells us that $c,d$ are integers and $c^2+d^2$ is odd and that $c+di$ is not divided by $1+i$.
Now, the common divisor of both $m+i$ and $m-i$ can divide its difference, and it is obvious that $2$ is not the divisor, so the common divisor are only units. Let's call this situation "$m+i$ and $m-i$ are coprime". Also, we see $n+i$ and $n-i$ are coprime. Of course, we see that $a+bi$ and $a-bi$ are coprime and that $c+di$ and $c-di$ are coprime. By substituting $m\pm i=(1\pm i)^{\mu}(a\pm bi)$ and $n\pm i=(1\pm i)^{\nu}(c\pm di)$ in $R$, we get $$(1-i)(1+i)^{\mu k+\nu l}(a+bi)^k(c+di)^l=(1+i)(1-i)^{\mu k+\nu l}(a-bi)^k(c-di)^l.$$ Noting that $1+i=i(1-i),$ then we get $$i^f(a+bi)^k(c+di)^l=(a-bi)^k(c-di)^l$$ where $f=\mu k+\nu l-1$. Since $a+bi$ and $a-bi$ are coprime, it must be that $$(a+bi)^k=\varepsilon (c-di)^l$$ where $\varepsilon$ is an unit. By the way, since $k$ and $l$ are coprime, we get $$a+bi=\varepsilon_1(\alpha +\beta i)^l, c-di=\varepsilon_2(\alpha+\beta i)^k$$ where $\varepsilon_1, \varepsilon_2$ are units and $\alpha, \beta$ are 'normal' integers. Hence, we get $$m+i=\varepsilon_1(1+i)^{\mu}(\alpha+\beta i)^l, n+i=\overline{\varepsilon_2}(1+i)^{\nu}(\alpha-\beta i)^k.$$ Then we get $n-i=\varepsilon_2(1-i)^{\nu}(\alpha+\beta i)^k$. By adding $m+i$ tells us that $m+n$ can be divided by $\alpha+\beta i$. Since $m+n$ is a real number, $m+n$ can be divided by $\alpha-\beta i$. Hence, we know that it can be divided by ${\alpha}^2+{\beta}^2$. Taking the norm of these equations, letting $A={\alpha}^2+{\beta}^2,$ we know that $$m^2+1=2^{\mu}A^l, n^2+1=2^{\nu}A^k$$ where $A\gt 1$ and $A$ is odd.
Hence, we now reach the diophantine equations $x^2+1=y^p$ or $x^2+1=2y^p\ \ \ (x\gt 0, y\gt 1).$
The followings about these equations are known :
1. $x^2+1=y^p$ has no solution when $p\gt 1$.
2. $x^2+1=2y^p$ has no solution when $p$ has an odd factor.
3. $x^2+1=2y^4\ (y\gt 1)$ has only one solution $(x,y)=(239,13)$.
From 2 and 3, we know that if $x^2+1=2y^p\ (y\gt 1)$, then $p=1,2,4$ since $13$ is not a square number. Here, we have the following five cases :
$$(1)\ \begin{cases} m^2+1=A \\ n^2+1=2A, \\ \end{cases}\ (2)\ \begin{cases} m^2+1=A \\ n^2+1=2A^2, \\ \end{cases}\ (3)\ \begin{cases} m^2+1=2A \\ n^2+1=2A^2, \\ \end{cases}$$$$(4)\ \begin{cases} m^2+1=A \\ n^2+1=2A^4, \\ \end{cases}\ (5)\ \begin{cases} m^2+1=2A \\ n^2+1=2A^4. \\ \end{cases}$$
Let's solve each case.
$(1)$ Since $k=l=1, \mu=0, \nu=1$, we get $$m+i=\varepsilon_1(\alpha+\beta i), n+i=\varepsilon_2(1+i)(\alpha-\beta i)=\varepsilon_3(1+i)(m-i).$$ Hence, $n+i=\varepsilon_3\left\{(m+1)+(m-1)i\right\}$. Since the imaginary part of the left side is $1$, we get $m\pm 1=\pm 1$. Since $m\not=0$, $m=\pm 2$. From its real part, we get $m\pm 1=\pm n.$ Since $|m|\lt |n|$, $n=\pm 3$. This leads $$\frac{\pi}{4}=\arctan\frac{1}{2}+\arctan\frac{1}{3}.$$
$(2)$ Since $k=1, l=2, \mu=0, \nu=1$, we get $$m+i=\varepsilon_1(\alpha+\beta i), n+i=\varepsilon_2(1+i)(\alpha-\beta i)^2=\varepsilon_4(1+i)(m-i)^2.$$ Hence, $n+i=\varepsilon_4\left\{(m^2+2m-1)+(m^2-2m-1)i\right\}$. We get $m^2\pm 2m-1=\pm 1$. Since $m$ is an integer, the right side$\not=+1$. Hence, $m(m\pm 2)=0$. Since $m\not=0$, $m=\pm 2$. Then, $n=\pm 7, A=5.$ Since $m+n$ can be divided by $5$, each sign of $m$ and $n$ is opposite. Then, we get $$\frac{\pi}{4}=2\arctan\frac{1}{2}-\arctan\frac{1}{7}.$$
$(3)$ Since $k=1, l=2, \mu=\nu=1$, we get $$m+i=\varepsilon_1(1+i)(\alpha+\beta i), n+i=\overline{\varepsilon_2}(1+i)(\alpha-\beta i)^2=\overline{\varepsilon_2}i(1-i)(\alpha-\beta i)^2.$$ Hence, $2n+2i=\varepsilon_5(1+i)(m-i)^2=\varepsilon_5\left\{(m^2+2m-1)+(m^2-2m-1)i\right\}$. So, we get $m^2\pm 2m-1=\pm 2$. Hence, $m=\pm 1, \pm 3$. Since ${\alpha}^2+{\beta}^2\gt 1$, $m\not=\pm 1$. So, $m=\pm 3.$ Then, $n=\pm 7, A=5.$ Since $m+n$ can be divided by $5$, the signs of $m$ and $n$ are the same. Then, we get $$\frac{\pi}{4}=2\arctan\frac{1}{3}+\arctan\frac{1}{7}.$$
In the cases of the both $(4)$ and $(5)$, we get $A=13, n=\pm 239$.
$(4)$ Since $m^2+1=13$, no solutions can be gotten.
$(5)$ Since $m^2+1=2\cdot 13$, we get $m=\pm 5$. Since $m+n$ can be divided by $13$, each sign of $m$ and $n$ is opposite. Then, we get $$\frac{\pi}{4}=4\arctan\frac{1}{5}-\arctan\frac{1}{239}.$$
Now the proof is completed.