Reciprocal Sum Equality

Without loss of generality, we may assume that $a \geq b$ . Note that as $c$ is positive, we have $\frac {1}{3} < \frac {1}{3} + \frac {1}{c+3} = \frac {1}{a+3} + \frac {1}{b+3} \leq \frac {2}{b+3}$ . This implies that $b+3 < 6 \Rightarrow b<3$ . We split into cases, depending on the value of $b$ .

Case 1: $b=1$ . The original equation becomes $\frac {1}{a+3} + \frac {1}{4} + \frac {1}{3} + \frac {1}{c+3}$ , which is equivalent to $\frac {1}{12} + \frac {1}{c+3} - \frac {1}{a+3} =0$ . Cross multiplying, we obtain $(a+3)(c+3) + 12 (a+3)-12(c+3) = 0$ . We 'complete the factorization' by subtracting 144 on both sides to obtain $(a-9)(c+15) = -144$ . Since $a$ and $c$ are positive integers, so $c + 15 > 0$ which implies that $a - 9 < 0$ , thus $a < 9$ .

We check all the possible values of $1 \leq a \leq 9$ , to get that the solutions are $(a,c) = (1,3), (3,9), (5, 21), (6, 33), (7, 57), (8,129)$ .

Case 2: $b=2$ . The original equation becomes $\frac {1}{a+3} + \frac {1}{5} = \frac {1}{3} + \frac {1}{c+3}$ . Similar as above, we cross multiply to obtain $2(a+3)(c+3) + 15(a+3) - 15 (c+3) = 0$ . This is harder to factorize using the previous trick directly, but we multiply by 2 and subtract 225 to obtain $(2a-9)(2c+21)=-225$ . Likewise, since $a$ and $c$ are positive integers, so $2c+21 > 0 \Rightarrow 2a-9<0 \Rightarrow a<4.5$ .

We check all the possible values of $1 \leq a \leq 4$ to get that the solutions are $(a, c) = (2, 12), (3, 27), (4, 102)$ .

Hence, the sum of all possible values of $c$ are $3 + 9 + 21 + 33+ 57 + 129 + 12 + 27 + 102 = 393$ .

Without loss of generality, we may assume that $a \leq b$ . Clearly $\mbox{RHS} > \frac{1}{3}$ , thus $\mbox{LHS} > \frac {1} {3}$ . If $a = 3$ , then $b \geq 3$ and $\frac {1} {6} + \frac {1} {b+3} \leq \frac {1}{3}$ . Hence $a = 1$ or $a= 2$ .

Case 1: $a = 1$ . $\begin{aligned} \frac {1}{b+3} & = \frac {1}{12} + \frac {1}{c+3} & \\ 12(c+3) & = (b+3)(c+3) + 12(b+3) & \\ bc+ 15b -9c + 9 & = 0 & \\ (b-9)(c+15) & = -144 = -12^2 & \\ \end{aligned}$

Since $c+15> 0$ thus we must have $b-9 <0$ , but $b \geq a = 1$ so $b -9 \geq -8$ . Hence we have $(b,c) = (1,3), (3,9), (5,21), (6,33),(7,57)$ and $(8,129)$ as solutions.

Case 2: $a = 2$ . $\begin{aligned} \frac {1} {b+3} & = \frac {2}{15} + \frac {1}{c+3} & \\ 15(c+3) & = 2(b+3)(c+3) + 15(b+3) & \\ 0 & = 2bc + 21b -9 c +18 & \\ 0 & = 4bc + 42 b - 18 c + 36 & \\ (2b - 9 )(2c + 21 ) & = -225 = -3^2\times 5^2 & \\ \end{aligned}$

Since $2c+21 > 0$ , we must have $2b-9< 0$ . But $b\geq a = 2$ , so $2b-9 \geq -5$ . Hence $2b - 9 = -5, -3, -1$ . This leads to corresponding solutions $(b,c) = (2, 12), (3, 27)$ and $(4, 102)$ .

Hence, the sum of all possible values of $c$ are $3 + 9 + 21 + 33+ 57 + 129 + 12 + 27 + 102 = 393$ .

$Let\quad a+3=A,\quad b+3=B,\quad c+3=C\quad just\quad to\quad make\quad the\quad \\ equations\quad look\quad neat.\\ \displaystyle \frac { 1 }{ a+3 } +\frac { 1 }{ b+3 } =\frac { 1 }{ 3 } +\frac { 1 }{ c+3 } \\ becomes\\ \displaystyle \frac { 1 }{ A } +\frac { 1 }{ B } =\frac { 1 }{ 3 } +\frac { 1 }{ C } \\ \displaystyle C=\frac { 1 }{ \frac { 1 }{ A } +\frac { 1 }{ B } -\frac { 1 }{ 3 } } \\ \displaystyle C=\frac { 3AB }{ 3(A+B)-AB } \\ To\quad satisfy\quad the\quad given\quad conditions,\\ 1.\quad A\ge 4\quad (Since\quad a\ge 1)\\ 2.\quad B\ge 4\quad (Since\quad b\ge 1)\\ And\quad the\quad denominator\quad has\quad to\quad be\quad positive,\quad \\ i.e,\quad 3(A+B)-AB>0\quad or\\ 3.\quad 3(A+B)>AB\\ Using\quad the\quad above\quad conditions\quad we\quad can\quad assign\quad \\ A\quad and\quad B\quad certain\quad values\quad to\quad see\quad which\quad all\quad \\ satisfy\quad the\quad conditions.\quad And\quad since\quad the\quad equation\quad is\quad \\ symmetric\quad w.r.t\quad A\quad and\quad B,\quad we\quad can\quad keep\quad A\quad \\ constant\quad and\quad vary\quad the\quad other\quad such\quad that\quad B\ge A.\\ Case\quad 1:\\ A=4.\\ We\quad see\quad the\quad only\quad values\quad of\quad B\ge A\quad \\ that\quad satisfy\quad the\quad condition\quad 3\quad and\quad 2\quad are:\\ B=4,5,6,7,8,9,10,11\\ Case\quad 2:\\ A=5.\\ We\quad see\quad the\quad only\quad values\quad of\quad B\ge A\quad \\ that\quad satisfy\quad the\quad condition\quad 3\quad and\quad 2\quad are:\\ B=5,6,7\\ Case\quad 3:\\ A=6\\ No\quad values\quad of\quad B\ge A\quad satisfy\quad the\quad give\quad conditions.\\ Last\quad condition:\\ C=Integer.\\ To\quad see\quad the\quad values\quad of\quad A\quad and\quad B\quad that\quad satisfy\quad that,\quad \\ I\quad just\quad plug\quad in\quad the\quad values\quad and\quad \\ discard\quad the\quad values\quad that\quad don't\quad satisfy.\\ We\quad find\quad out,\\ C=6,12,24,36,60,132,15,30,105\\ Since\quad C=c+3,c=C-3\\ c=3,9,21,33,57,129,12,27,102\\ Summing\quad them\quad up,\\ Answer=\boxed { \boxed { 393 } } \\$