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#1 2018-08-10 02:32:19

KerimF
Member
From: Aleppo-Syria
Registered: 2018-08-10
Posts: 8

Playing with natural numbers and SQRT

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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#2 2018-08-10 05:18:54

Alg Num Theory
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Registered: 2017-11-24
Posts: 332
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Re: Playing with natural numbers and SQRT

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

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#3 2018-08-10 09:44:18

KerimF
Member
From: Aleppo-Syria
Registered: 2018-08-10
Posts: 8

Re: Playing with natural numbers and SQRT

Alg Num Theory wrote:

If a = 2, b = 1, c = 4, then A = √(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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