Given a number x, are there two numbers, a and b, where a+b=x AND a*b=x?

[bgold024]

I was just wondering if it is already known which two numbers can both add AND multiply to equal an arbitrary number?

For example, if I choose any number, x, can you tell me what two numbers can both add AND multiply to equal x?

A very easy x would be 4.

2+2=4

2×2=4

In that case, a and b are the same number, 2, but what if I told you that x has to be included.  Can you satisfy the following equality:

4+4=4

AND

4×4=4

Obviously those are not true, but what transform could you do to the left hand sides, without changing the right hand sides.  Simply put, how could you make 4+4 turn into 2+2 and 4×4 turn into 2×2, using the number 4 twice with the same operations, and make them both equal to 4.

(Even harder)

What if x was √2, or π, or i, or anything?  How do you make:

√2+√2=√2 AND √2×√2=√2

Is it already known which two numbers can both sum AND produce any given x?

I found a very cool equality, where using the arbitrary x yeilds both numbers which satisfy the above and was wondering if anyone else found it and if it means anything?

One last way I guess I can say it is, what transformation will make x^2/2x=1?

(keep in mind both x^2 and 2x will become the transform used to make a+b=a×b, where a and b are relatively the same)

The solution is very similar to another famous formula, and also has an interesting, but expected result.  Honestly the solve for this is quite trivial, but it just works out so nicely

 

[Strange]

Your post seems to talk about a special case, where a=b.

In this case it is trivial to see that x^2 = 2x, therefore x = 2. Which is the first example you started with. And the only example with a = b.

Looking at the question in the title: are there two numbers, a and b, where a+b=x AND a*b=x?

That is obviously not true for all x. For example, if x is prime.

In other words, you want to find a and b, such that: (a+b) = ab

I think a=b=0 and a=b=2 are the only integer solutions. 

Finding non-integer solutions is trivial. We can rewrite the above as a = b/(b-1).

Then set b=5, for example and we get a = 1.25 (5+1.25 = 6.25 and 5*1.25 = 6.25)

[bgold024]

@Strange You are both correct in a non integer solution as b +× b/(b-1) and the only integer solution being 2 and 0, but you can absolutely do this for every number (prime, transcendental, real, imaginary.... Anything.)  There isn't a restriction on the set of numbers you can use, and I suppose you could call this a special case.

Your second solution is a little closer, but there's an equality very similar to a famous equation as the solution.

I guess another way to put it, take a number, x, and split it into two parts such that both parts can sum and produce x.

Pretty sure it's not known, I haven't seen anything like it.  Probably thinking it's kinda useless, but might be a cool design technique for layouts.

Edited October 1, 2019 by bgold024

[Strange]

1 hour ago, bgold024 said:

but you can absolutely do this for every number (prime, transcendental, real, imaginary

You obviously can't do it for primes, by definition.

[uncool]

3 hours ago, Strange said:

You obviously can't do it for primes, by definition.

You seem to be assuming only integer solutions are acceptable, which seems unwarranted.

You can find a real a and b as long as x is at least 4 or negative.

[Strange]

4 minutes ago, uncool said:

You seem to be assuming only integer solutions are acceptable, which seems unwarranted.

Not at all. I demonstrated that there are an infinite number of non-integer solutions. 

I was just correcting your claim that it applies to primes. And I’m pretty sure there are no other integer solutions than 0 or 4.

*thinks...*

Are you suggesting there are non-integer a and b that produce a prime x? I suppose that is possible. Do you have an example?

*thinking...*

No, I don’t think it is possible (but can’t prove it)

*thinks some more*

Of course it is possible: b^2 -xb + x = 0