The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

Let's assume the three numbers are a, b, and c.

According to the given information, we have two equations:

1. a^2 + b^2 + c^2 = 138
2. ab + ac + bc = 131

To find the sum of the three numbers, we need to solve these equations simultaneously.

First, let's square equation 2:

(ab + ac + bc)^2 = 131^2 a^2b^2 + a^2c^2 + b^2c^2 + 2ab^2c + 2abc^2 + 2a^2bc = 17161

Now, let's subtract equation 1 from this equation:

(a^2b^2 + a^2c^2 + b^2c^2 + 2ab^2c + 2abc^2 + 2a^2bc) - (a^2 + b^2 + c^2) = 17161 - 138

Simplifying this equation, we get:

a^2b^2 + a^2c^2 + b^2c^2 + 2ab^2c + 2abc^2 + 2a^2bc - a^2 - b^2 - c^2 = 17023

Rearranging the terms, we have:

a^2(b^2 + c^2 - 1) + b^2(c^2 + a^2 - 1) + c^2(a^2 + b^2 - 1) + 2abc(ab + ac + bc) = 17023

Since we know that ab + ac + bc = 131, we can substitute this value into the equation:

a^2(b^2 + c^2 - 1) + b^2(c^2 + a^2 - 1) + c^2(a^2 + b^2 - 1) + 2abc(131) = 17023

Simplifying further, we get:

a^2(b^2 + c^2 - 1) + b^2(c^2 + a^2 - 1) + c^2(a^2 + b^2 - 1) + 262abc = 17023

Now, let's substitute the value of a^2 + b^2 + c^2 from equation 1 into this equation:

138(b^2 + c^2 - 1) + 138(b^2 + c^2 - 1) + 138(b^2 + c^2 - 1) + 262abc = 17023

Simplifying further, we get:

414(b^2 + c^2 - 1) + 262abc = 17023

Dividing both sides of the equation by 2, we have:

207(b^2 + c^2 - 1) + 131abc = 8511

Now, let's consider the possible values of b^2 + c^2 - 1. Since the sum of squares of three numbers is 138, we can write:

b^2 + c^2 = 138 - a^2

Substituting this value into the equation, we get:

207(138 - a^2 - 1) + 131abc = 8511

Simplifying further, we have:

207(137 - a^2) + 131abc = 8511

Expanding the equation, we get:

28359 - 207a^2 + 131abc = 8511

Rearranging the terms, we have:

131abc - 207a^2 = 8511 - 28359

Simplifying, we get:

131abc - 207a^2 = -19848

Now, let's consider the possible values of abc. Since the sum of products taken two at a time is 131, we can write:

ab + ac + bc = 131

Expanding this equation, we get:

ab + ac + bc - abc = 131

Rearranging the terms, we have:

abc - ab - ac - bc = -131

Substituting this value into the previous equation, we get:

(abc - ab - ac - bc) - 207a^2 = -19848

Simplifying further, we have:

abc - ab - ac - bc - 207a^2 = -19848

Now, let's factor out the common terms:

a(bc - b - c - 207a) = -19848

Dividing both sides of the equation by a, we have:

bc - b - c - 207a = -19848/a

Since a, b, and c are integers, the right side of the equation must also be an integer. Therefore, -19848/a must be an integer.

Now, let's consider the factors of 19848:

1, 2, 3, 4, 6, 8, 12, 16, 24, 41, 48, 82, 123, 164, 246, 328, 492, 656, 984, 1649, 1968, 3298, 4947, 6596, 9894, 13192, 19788

We can see that the only factor of 19848 that can make -19848/a an integer is 1. Therefore, a must be 1.

Substituting a = 1 into the equation, we have:

bc - b - c - 207(1) = -19848/1

Simplifying, we get:

bc - b - c - 207 = -19848

Rearranging the terms, we have:

bc - b - c = -19848 + 207

Simplifying further, we get:

bc - b - c = -19641

Now, let's factor out the common terms:

b(c - 1) - (c - 1) = -19641

Simplifying, we have:

(c - 1)(b - 1) = -19641

Since a, b, and c are integers, the factors on the left side of the equation must also be integers. Therefore, -19641 must have integer factors.

Now, let's consider the factors of 19641:

1, 3, 7, 9, 21, 27, 63, 189, 104, 312, 728, 936, 2184, 2808, 6552, 19656

We can see that the only factor of 19641 that can make (c - 1)(b - 1) an integer is 1. Therefore, (c - 1)(b - 1) must be 1.

Since 1 is a prime number, the only possible values for (c - 1) and (b - 1) are 1 and -1, or -1 and 1.

Case 1: (c - 1) = 1 and (b - 1) = 1 Solving these equations, we get c = 2 and b = 2.

Case 2: (c - 1) = -1 and (b - 1) = 1 Solving these equations, we get c = 0 and b = 2.

Now, let's substitute these values into the equation bc - b - c = -19641:

Case 1: (2)(2) - 2 - 2 = -19641 Simplifying, we get 4 - 4 = -19641, which is not true.

Case 2: (0)(2) - 2 - 0 = -19641 Simplifying, we get -2 - 0 = -19641, which is not true.

Since neither case satisfies the equation, there are no integer solutions for a, b, and c.

Therefore, it is not possible to determine the sum of the three numbers.