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| Some Reals Sum Integer (Posted on 2008-08-07) |
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There are precisely two nonnegative real values of R, so that for each of these values:
3√(3 + √R) + 3√(3 - √R) is an integer.
Find these values, and prove that no other nonnegative real R can conform to the given conditions.
*** While the solution may be facile with the aid of a computer program, show how to derive it without one.
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Submitted by K Sengupta
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Rating: 5.0000 (1 votes)
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Solution:
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R = 368/27, 242/27 are the required possible nonnegative values of R.
EXPLANATION:
Daniel has submitted a methodology here .
Daniel has also sumitted a second methodology, which matches with the one known to me, and the said procedure is reproduced in full.
At the outset, we observe that the given expression is not an integer for R=0.
For R > 0, let M = (3 + √R)1/3 + (3 - √R)1/3
Then, M3 = 6 + 3*(3 + √R)1/3*(3 - √R)1/3*M
Upon due simplification, we have:
(M3 - 6)3/(3M)3 = 9 – R, so that:
R = 9 - (M3 - 6)3/(3M)3 > 0 …..(i)
Since, (M2/3 - 2/M)3 is monotone increasing and > 9 for M ≥ 3, it follows that M = 1, 2
Substituting M =1 in (i), we obtain R = 368/27, and for M=2, we have:
R = 242/27
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