Pythagorean computation

The key to this problem is algebra. Take the sum of the three squares 2^2+3^2+6^2. This is equal to 7^2. We can show algebraically that x^2+(x+1)^2+(x^2+x)^2=x^2+x^2+2x+1+x^4+2x^3+x^2=x^4+2x^3+3x^2+2x+1=(x^2+x+1)^2. Now notice that 2(10^102)+2(10^51)+(10^102+10^51)^2+1=10^102+2 10^51+1+10^102+((10^51)(10^51+1))^2. Since x^2+2x+1=(x+1)^2, 10^102+2 10^51+1=(10^51+1)^2. So 10^102+2*10^51+1+10^102+((10^51)(10^51+1))^2=(10^51+1)^2+(10^51)^2+((10^51)(10^51+1))^2. Notice that this is in the form of x^2+(x+1)^2+(x(x+1))^2. So this is equal to (x^2+x+1)^2=(10^102+10^51+1)^2. so the square root of this quantity is 10^102+10^51+1. This is equal to 10^(102)+10^(102/2)+1. So A is equal to 102. The prime factorization of 102 is 2 *3 *17. This implies that the sum of the prime factors of A is 2+3+17= 22 .

Set $x=10^{51}$ . Then $\begin{aligned} 2\cdot10^{102} + 2\cdot 10^{51} + (10^{102} + 10^{51})^2 + 1 &= 2x^2 + 2x + (x^2 + x)^2 + 1 \\ &= (x^2 + x)^2 + 2(x^2 + x) + 1 \\ &= [(x^2 + x) + 1]^2 \end{aligned}$ Thus, the square root of the given expression is simply $10^{102} + 10^{51} + 1$ and $A = 102 = 2\times 3\times 17$ . $2 + 3 + 17 = \boxed{22}$ .