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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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See also math.stackexchange.com/questions/44595/…. –  lhf Sep 30 '13 at 19:00

2 Answers 2

up vote 2 down vote accepted

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.

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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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