# $\frac{\pi}{4}=k\arctan \frac{1}{m}+l\arctan \frac{1}{n}$ has only four solutions?

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?

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– lhf Sep 30 '13 at 19:00

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}.$$Now the proof is completed. - It is not answer but I wanted to figure out an point about 4 solutions$$\frac{\pi}{4}=\arctan \frac{1}{2}+\arctan \frac{1}{3}Z_1=(2+i)(3+i)=-1+6 +i(2+3)=+5+5i\frac{\pi}{4}=2\arctan \frac{1}{2}-\arctan \frac{1}{7}Z_2=(2+i)^2 (7+i)^{-1}=\frac{3+4i}{7+i}=\frac{(3+4i)(7-i)}{50}=\frac{25}{50}+i\frac{25}{50}$$If you also write other 2 examples as a complex number, you will notice that same thing. real and imaginary parts are equal because of \frac{\pi}{4}=\arctan 1. Thus your question is now to find complex numbers that$$\frac{\pi}{4}=k\arctan \frac{1}{m}+l\arctan \frac{1}{n}\tag{$\star$}Z=(m+i)^k (n+i)^{l}=a(1+i)$\$ I believe that to focus in such way can be more helpful for your research. If I find any further step to find other examples I will let you know

EDIT: Check here for more information. There is a note in wiki page that "It is believed that there are exactly four solutions". But there is no proof .Seems that your question is a open problem from past.

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