C# Make a sum with two random numbers but it must equal the users input

Question

I was wondering if it is possible to make a sum from a user input int but the sum must be radom every time.

This is the code that I made :

    public static string genSum(int askedNumber)
    {
        string outputStr = null;

        bool switchOneTwo = false;
        int neededSum = 0;

        int intOne = 0;
        int intTwo = 0;

        while (askedNumber != neededSum)
        {

            if (switchOneTwo == true)
            {
                intOne += 1;
                switchOneTwo = false;
            }
            else
            {
                intTwo += 1;
                switchOneTwo = true;
            }

            neededSum = intOne + intTwo;

            if (neededSum == askedNumber)
            {
                if (neededSum >= 4)
                {
                    Random randomInt = new Random();

                    int tmpIntOne = intOne;
                    int tmpIntTwo = intTwo;
                    int method;

                    //For now only one method
                    method = randomInt.Next(1, 1);

                    if (method == 1)
                    {
                        tmpIntOne = tmpIntOne / 2;
                        tmpIntTwo = (tmpIntTwo * 2) - tmpIntOne;
                    }

                    intOne = tmpIntOne;
                    intTwo = tmpIntTwo;                
                }

                outputStr = "(" + intOne.ToString() + "+" + intTwo.ToString() + ")";

                return outputStr;
            }
        }
        return outputStr;
    }

so what I want is that if a user eneters a number for example 20 it will then make a sum :

User enters 20 and presses GO! :

Result :

10 + 10 = 20

12 + 8 = 20

5 + 15 = 20

1 + 19 = 20

3 + 17 = 20

User enters 500 and presses GO! :

Result :

9 + 491 = 500

263 + 237 = 500

300 + 200 = 500

250 + 250 = 500

109 + 391 = 500

Answer 1

Create a random number less than the input and the other one will be User-YourRandom

void Foo(int userNumber)
{
    Random r = new Random();
    int firstNumber = r.Next(userNumber - 1);
    int secondNumber = userNumber - firstNumber;
}

You can extend this to support negatives also.

edit getting true randoms

as I4V points out, if you put this in a loop you might get same set of numbers, so instead of calling Random you can use this approach:

void Foo(int userNumber)
{
    int userNumber = 500;

    for (int i = 0; i < 10; i++ )
    {
        int firstNumber = Next(0, userNumber - 1);
        int secondNumber = userNumber - firstNumber;

        Console.WriteLine(firstNumber + "+" + secondNumber);
    }
}

internal static int RandomExt(int min, int max)
{
    RNGCryptoServiceProvider rng = new RNGCryptoServiceProvider();
    byte[] buffer = new byte[4];

    rng.GetBytes(buffer);
    int result = BitConverter.ToInt32(buffer, 0);

    return new Random(result).Next(min, max);
}

Answer 2

foreach (var s in genSum(500).Take(5))
{
    Console.WriteLine(s);
}

---

public static IEnumerable<string> genSum(int askedNumber)
{
    Random r = new Random();
    while (true)
    {
        var i = r.Next(0, askedNumber);
        yield return i + "+" + (askedNumber - i);
    }
}