Sum of 10's a pie

Let $a=3+i$ , $b=4+i$ , $c=5+i$ and $d=n+i$ . So,the expression we have is the sum of the arguments of the complex numbers above. But also, this sum is the argument of the product of these complex numbers, $abcd$ , so:

$abcd=(3+i)(4+i)(5+i)(n+i)=(11+7i)(5n-1+i(n+5)) \\ abcd=48n-46+i(46n+48)$

Now, taking the argument of $abcd$ we have:

$\arctan\left(\dfrac{46n+48}{48n-46}\right)=\dfrac{\pi}{4} \\ \dfrac{46n+48}{48n-46}=1 \\ 46n+48=48n-46 \\ 94=2n \\ n=\boxed{47}$

Let $\alpha = \arctan {\frac{1}{3}}$ , $\beta = \arctan {\frac{1}{4}}$ , $\gamma = \arctan {\frac{1}{5}}$ and $\delta = \arctan {\frac{1}{n}}$ . Then,

$\begin{aligned} \alpha + \beta + \gamma + \delta & = \dfrac{\pi}{4} \\ \tan{(\alpha + \beta + \gamma + \delta)} & = \tan{\dfrac{\pi}{4}} \\ \frac {\tan{(\alpha + \beta)} + \tan {(\gamma + \delta)}} {1 - \tan{(\alpha + \beta)} \tan {(\gamma + \delta)}} & = 1 \\ \frac {\frac{\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{12}} + \frac{\frac{1}{5}+\frac{1}{n}}{1-\frac{1}{5n}}} {1 - \frac{\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{12}} \times \frac{\frac{1}{5}+\frac{1}{n}}{1-\frac{1}{5n}}} & = 1 \\ \frac {\frac{7}{11}+\frac{n+5}{5n-1}} {1-\frac{7}{11} \times \frac{n+5}{5n-1}} & = 1 \\ \frac {35n-7+11n+55} {55n-11-7n-35} & = 1 \\ \frac {46n+48} {48n-46} & = 1 \\ 46n + 48 & = 48n - 46 \\ 2n & = 94 \\ n & = \boxed{47} \end{aligned}$

$\arctan\left(\frac {1}{3} \right)+ \arctan\left(\frac {1}{4}\right) + \arctan\left(\frac {1}{5} \right)- \arctan1=-\arctan\left(\frac {1}{n} \right)$

$\arctan\left(\frac {\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{3}\times{\frac{1}{4}}}\right)+\arctan\left(\frac {\frac{1}{5}-1}{1+\frac{1}{5}\times{1}}\right)=-\arctan\left(\frac {1}{n} \right)$

$\arctan\left(\frac {7}{11} \right)+ \arctan\left(-\frac {2}{3}\right)=-\arctan\left(\frac {1}{n} \right)$

$\arctan\left(\frac {\frac{7}{11}-\frac{2}{3}}{1+\frac{7}{11}\times{\frac{2}{3}}}\right)=-\arctan\left(\frac {1}{n} \right)$

$\arctan\left(-\frac {1}{47}\right)=-\arctan\left(\frac {1}{47}\right)=-\arctan\left(\frac {1}{n} \right)$

$n=47$