any numbersThe result=1 and the sum =1How to find these numbers?

Question

Problem: There are 4 any numbers: a, b, c, and d. The result of multiplication of these number is 1 (a*b*c*d=1). The sum of these numbers is also 1 (a+b+c+d=1).How to find these numbers? Does the solution of this problem exist? If not, why?

Answer 1

There are lots of answers, none with integer answers, but if you are allowed real numbers you can solve this...

One way to do it would be to break the equations into two parts:

Let's say we could figure a way to make:
a * b = -1
a + b = 1/2

And do the same for c and d:
c * d = -1
c + d = 1/2

Then you'd have to agree that:
(a * b) * (c * d) = -1 * -1 = 1
(a + b) + (c + d) = 1/2 + 1/2 = 1

Right?

So let's start with those equation:
a * b = -1
a + b = 1/2

From the first equation, solve for a:
a = -1/b

Substitute into the second equation:
-1/b + b = 1/2

Multiply by b on both sides (obviously b can't be zero):
-1 + b^2 = b/2

Multiply by 2 on both sides:
-2 + 2b^2 = b

Subtract b from both sides:
2b^2 - b - 2 = 0

Now use the quadratic formula to figure the roots:
You get:
(1 + sqrt(17)) / 4 ≈ 1.280776406...
(1 - sqrt(17)) / 4 ≈ -0.780776406...

Set a to one root and b to the other.
a = 1.280776406...
b = -0.780776406...

Just to prove that this works:
a + b = (1 + sqrt(17)) / 4 + (1 - sqrt(17)) / 4
a + b = (1 + sqrt(17) + 1 - sqrt(17)) / 4
a + b = (1 + 1) / 4
a + b = 2 / 4
a + b = 1/2
c + d = 1/2
a + b + c + d = 1

a * b = (1 + sqrt(17)) / 4 * (1 - sqrt(17)) / 4
a * b = (1 + sqrt(17))(1 - sqrt(17)) / 16
a * b = (1 + sqrt(17) - sqrt(17) - sqrt(17)^2) / 16
a * b = (1 - sqrt(17)^2 ) / 16
a * b = (1 - 17) / 16
a * b = -16 / 16
a * b = -1
c * d = -1
a * b * c * d = 1

So one answer is:
a = (1 + sqrt(17)) / 4 ≈ 1.280776406...
b = (1 - sqrt(17)) / 4 ≈ -0.780776406...
c = (1 + sqrt(17)) / 4 ≈ 1.280776406...
d = (1 - sqrt(17)) / 4 ≈ -0.780776406...

Answer 2

bandf has given a very good solution which can be generalised a bit.

We could let the numbers be a, 1/a, b, 1/b, and find the condition for their sum to be 1. There are plenty of other solutions, but this is an approach which includes bandf's solution.

We can't just put in any value for 'a', because that won't always give a real value for b. Let k be the value of a +1/a. Then we need to solve
b + 1/b = 1-k

i.e. b^2 -(1-k)b + 1 = 0

This equation has real roots only if
(1-k)^2 - 4 > 0 (or equal)

k^2 - 2k - 3 > 0
and so k < -1 or k > 3

So we can pick any value of 'a' which makes
a + 1/a less than -1 or greater than 3.

e.g. We could choose a = 3, and then solve
3 + 1/3 + b + 1/b = 1
b + 7/3 + 1/b = 0

3b^2 + 7b + 3 = 0

One of the solutions is (-7 + sqrt(13));
the other is the value of 1/b, i.e. (-7 - sqrt(13))

Hence 3, 1/3, -7 + sqrt(13), -7 - sqrt(13) is another of the "zillions" of solutions to this problem.

Yet another approach would be to let the numbers be
a, b, x, 1/(abx) and then find the condition that their sum is 1.

If you look at the condition as a quadratic equation in x, it turns out that for real roots we need
a^2 b^2 (a + b - 1)^2 > 4

One such pair of values is a = 2, b = 1, and if you solve

2 + 1 + x + 1/2x = 1

you get x = (-2 + sqrt(2))/2 as one of the solutions.

Answer 3

Probably zillions of solutions. You have two equations and four unknowns. The solutions may not be rational, and might not be easy to find.