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**KerimF****Member**- From: Aleppo-Syria
- Registered: 2018-08-10
- Posts: 14

We have the two algebraic expressions A and B:

A= SQRT[ a + b*SQRT(c) ] and B= SQRT[ a - b*SQRT(c) ]

If “a”, “b” and “c” are natural numbers (positive integers), A and B will likely be real numbers, not natural ones.

But if their sum A+B=S comes out as a natural number, S equals “b” always.

First, I wonder if there is even one set of “a”, “b and “c” (natural numbers) that lets S be a natural number as well but not equal to “b”.

I personally couldn't find it.

Second, in case S=b (as it is supposed to be), the 3 numbers satisfy the relation c = a – b^2/4.

For example, if a=19 and b=8, c=3 (and S=b=8).

This is an example about how the sum of two non-finite decimal numbers (as A and B here) could be a nice integer.

Kerim

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**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 693
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If *a* = 2, *b* = 1, *c* = 4, then *A* = √(2+1·√ 4) = 2, *B* = √(2−1·√ 4) = 0, *S* = *A* + *B* = 2 ≠ *b*.

Me, or the ugly man, whatever (3,3,6)

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**KerimF****Member**- From: Aleppo-Syria
- Registered: 2018-08-10
- Posts: 14

Alg Num Theory wrote:

If

a= 2,b= 1,c= 4, thenA= √(2+1·√ 4) = 2,B= √(2−1·√ 4) = 0,S=A+B= 2 ≠b.

Good work.

In this case, the original statement should be updated as:

"If their sum *A+B=S* comes out as a natural number, *S* equals *b* if *b* ≠ 1".

Thank you.

Added:

also *a* ≠ 0, *b* ≠ 0 and *c* ≠ 0

*Last edited by KerimF (2018-08-10 23:02:31)*

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