Square-Root Equation

$\begin{aligned} \frac 1{\sqrt{x^2+2x+2}} + \frac 1{\sqrt{3x^2-2x+1}} & = \frac 1{\sqrt{2x^2+x+2}} + \frac 1{\sqrt{2x^2-x+1}} \\ \frac 1{\sqrt{x^2+2x+2}} + \frac 1{\sqrt{2x^2-x+1+\color{#D61F06}x^2-x}} & = \frac 1{\sqrt{x^2+2x+2 +\color{#D61F06} x^2-x}} + \frac 1{\sqrt{2x^2-x+1}} \end{aligned}$

We note that if $\color{#D61F06}x^2 - x =0$ , then $LHS = RHS$ . Therefore, $x=0$ and $x=1$ are the integer roots of the equation, since $x^2+2x+2 \ne 0$ and $2x^2 -x+1\ne 0$ when $x=0, 1$ .

We also note that $LHS=RHS$ when $x^2+2x+2 = 2x^2 -x+1$ . Then $x^2 - 3x - 1 =0$ . Solving the quadratic equation we have $x = \dfrac {3\pm \sqrt{13}}2$ .

Therefore, $m+n+a+b+c = 0+1+3+13+2 = \boxed{19}$ .

Let $P = \sqrt{x^{2}+2x+2}$ ,

$Q = \sqrt{3x^{2}-2x+1}$ ,

$R = \sqrt{2x^{2}+x+2}$ ,

$S = \sqrt{2x^{2}-x+1}$ .

Note that $P^{2}+Q^{2} = R^{2}+S^{2}$ .

The equation is $\dfrac{1}{P}+\dfrac{1}{Q} = \dfrac{1}{R}+\dfrac{1}{S}$ .

$\dfrac{(P+Q)^{2}}{P^{2}Q^{2}} = \dfrac{(R+S)^{2}}{R^{2}S^{2}}$

$(P^{2}+Q^{2}+2PQ)(R^{2}S^{2}) = (R^{2}+S^{2}+2RS)(P^{2}Q^{2})$

$(P^{2}+Q^{2}+2PQ)(R^{2}S^{2}) = (P^{2}+Q^{2}+2RS)(P^{2}Q^{2})$

$(P^{2}+Q^{2})(R^{2}S^{2}-P^{2}Q^{2})+2PQRS(RS-PQ) = 0$

$(P^{2}+Q^{2})(RS-PQ)(RS+PQ)+2PQRS(RS-PQ) = 0$

$(RS-PQ) \{ (P^{2}+Q^{2})(RS+PQ)+2PQRS \} = 0$

Since $P,Q,R,S > 0$ , it follows that $(P^{2}+Q^{2})(RS+PQ)+2PQRS > 0$ .

Therefore, $RS = PQ$ .

$\sqrt{2x^{2}+x+2}\sqrt{2x^{2}-x+1} = \sqrt{x^{2}+2x+2}\sqrt{3x^{2}-2x+1}$

$4x^{4}+5x^{2}-x+2 = 3x^{4}+4x^{3}+3x^{2}-2x+2$

$x^{4}-4x^{3}+2x^{2}+x = 0$

$x(x-1)(x^{2}-3x-1) = 0$

The two integer roots are $x = 0,1$ .

The two irrational roots are $x = \dfrac{3 \pm \sqrt{13}}{2}$ .

Therefore, $m+n+a+b+c = 0+1+3+13+2 = 19$ .