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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
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
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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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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)
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
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