Sum of arctangents equals to pi/4

Suppose $a\ge 4$ . Then $\frac{1}{4}\ge \frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}$ .

Then $3\arctan\left(\frac{1}{4}\right)\ge \arctan\left(\frac{1}{a}\right)+\arctan\left(\frac{1}{b}\right)+\arctan\left(\frac{1}{c}\right)=\frac{\pi}{4}$

This is impossible as $\tan\frac{\pi}{12}=2-\sqrt{3}>\frac{1}{4}$ .

Hence $a=2$ or $a=3$ .

Using the formula $\arctan x+\arctan y = \arctan \frac{ x + y}{1- xy}$ repeatly, we have $\arctan\left(\frac{ab+bc+ca-1}{abc-a-b-c}\right)=\frac{\pi}{4}$ . This means that $ab+bc+ca-1=abc-a-b-c$ . We can rewrite it as

$c=\frac{(a+1)(b+1)-2}{(a-1)(b-1)-2}$

If $a=2$ , then $c=\frac{3b+1}{b-3}=3+\frac{10}{b-3}$ . The possible values of $(a,b,c)$ are $(2,4,13)$ and $(2,5,8)$ .

If $a=3$ , then $c=\frac{2b+1}{b-2}=2+\frac{5}{b-2}$ . The possible value of $(a,b,c)$ is $(3,3,7)$ .

As such, $P=2(4)(13)+2(5)(8)+3(3)(7)=247$ .

From $\tan(\alpha + \beta + \gamma) = \frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\beta\tan\gamma-\tan\gamma\tan\alpha},$ it follows that the equation is satisfied if $1 = \frac{\tfrac1a+\tfrac1b+\tfrac1c-\tfrac1a\tfrac1b\tfrac1c}{1-\tfrac1a\tfrac1b-\tfrac1b\tfrac1c-\tfrac1c\tfrac1a};$ we multiply denominator and numerator with $abc$ and equate them: $ab + bc + ca - 1 = abc - a - b - c.$ At least one of $a, b, c$ should be fairly small; otherwise $abc$ becomes much greater than any of the terms on the left. We will therefore go through all possible values of $a$ , and solve for $(b, c)$ . Thus we write the equation as $(a-1)bc - (a+1)(b+c) - (a-1) = 0, \\ bc - \frac{a+1}{a-1}(b+c) - 1 = 0, \\ \left(b - \frac{a+1}{a-1}\right)\left(c - \frac{a+1}{a-1}\right) = 1+\left(\frac{a+1}{a-1}\right)^2.$ The solutions can only be integers if $a-1|a+1$ , which requires that $a-1|2$ , or $a = 2, 3$ .

This limits the possibilities severely! $\begin{array}{lllll} \hline a = 2 & (b-3)(c-3) = 10 & = 2\cdot 5 & b = 5,\ c = 8 & 80 \\ & & = 1\cdot 10 & b = 4,\ c = 13 & 104 \\ a = 3 & (b-2)(c-2) = 5 & = 1\cdot 5 & b = 3,\ c = 7 & 63 \\ \hline \end{array}$ The total of possible values of $abc$ is therefore $\boxed{247}$ .

I solved it with:

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var piOverFour=Math.PI/4;
var results=[];
for(var a=1;a<100;a++)for(var b=1;b<100;b++)for(var c=1;c<100;c++)
{
    if(a>b||a>c||b>c)continue;
    var r=Math.atan(1/a)+Math.atan(1/b)+Math.atan(1/c);
    if(r+0.00000001>piOverFour&&r-0.00000001<piOverFour)//allow for some JavaScript margin of error
    {
        console.log([a,b,c]);
        results.push([a,b,c]);
    }
}
var result=0;
for(var i=0;i<results.length;i++)
{
    var r=results[i][0]*results[i][1]*results[i][2];
    console.log(r);
    result+=r;
}
console.log('result: '+result);

Using calculator, we have:-
$Arctan(1/2)=26.56505^o...............Arctan(1/3)=18.43494^o.............Arctan(1/4)=14.03624^o\\ But~45/3=15^o.....\implies~~a=2,~0r~a=3,~ only,~ and~ b\neq 2.\\ \implies~One~ of~ the~ angle~ must ~be ~26.56505^o...~ OR~ ...18.43494^o,\\ Note:- Arctan(1/2)+Arctan(1/3)= 45^o !!!\\$ $Using~a=2 ~next~b=4,~~ 45 - Arctan(1/2) - Arctan(1/4) =4.398705355.~~~\dfrac 1{Tan4.398705355}=13. \\ \text{So the first set is } P_1=2*4*13=\color{#3D99F6}{104}.\\$ $Next~ in~ line~ is~b=5~~~~45 - Arctan(1/2)- Arctan(1/5)=7.125016349.~~~~~\dfrac 1 {Tan7.125016349}=8.\\ \text{So the second set is } P_2=2*5*8=\color{#3D99F6}{80}.\\$ $Next~ in~ line~ is~c=\dfrac 1 {Tan\{45 - Arctan(1/2)- Arctan(1/6)\}}=6\frac1 3.\\ \text{c can not be 6 nor 7. So we have come to the limit for a=2.} \\ Next,~a=3~and~ b=3,~~~c=\dfrac 1 {Tan\{45 - 2* Arctan(1/3)\}}= 7\\ \text{The third set is } P_3=3*3*7=\color{#3D99F6}{63}.\\ a=3~and~next~ b=4,~~~c=\dfrac 1 {Tan\{45 - Arctan(1/3) - Arctan(1/4)\}}= 4\frac 1 2. \\ \text{c can not be 4 nor 5. So we have come to the limit for a=3.} \\ P_1+P_2+P_3=104+80+63=\Large~~~~~\color{#D61F06}{147}.$