4 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

You said a, b, c and d were *numbers*, not integers. So there is a possibility for an answer:

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 done a very good solution. You could generalise it a bit by taking the numbers as a, 1/a, b, 1/b (yes, of course there are still other possibilities), but if you just put in any value for 'a' there won't always be a real solution for 'b'.

In fact, if a + 1/a = s, then to find b we solve

b + 1/b = 1 - s
i.e. b^2 - (1 - s)b + 1 = 0, which has real roots only if

(1-s)^2 - 4 > 0 (or equal)
so s^2 - 2s - 3 > 0
hence s > 3 or s < -1

For example, we could pick a =3

Then a + 1/a = 10/3, and b is a root of
3b^2 + 7b + 3 = 0

b = (-7 + sqrt(13))/6 and 1/b = (-7 - sqrt(13))/6,

and these two numbers together with 3 and 1/3
satisfy the condition.

Anyone coming up with an even more general solution?

Answer 3

no solution

because it has to be 1*1*1*1 to get the product of 1 (multiplaction) and 1+1+1+1= 4 and that doesn't equal 1 so that would make it no solution