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 Sum Reals, Get Integer (Posted on 2016-07-09)
Find all nonnegative real numbers R such that:

3√(13+√R) + 3√(13-√R) is an integer.

Prove that there are no others.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution | Comment 1 of 3
Let I = cbrt(13+sqrt(R)) + cbrt(13-sqrt(R))

If 0<R<=169 then both terms of I are non-negative.  If R>169 then the positive term will be greater than the absolute value of the negative term, so the sum will be positive.  This means I will always be positive.

Then express I^3 = 26 + 3*[cbrt(13+sqrt(R)) + cbrt(13-sqrt(R))]*[cbrt(169-R)].

Substitute the original expression for I and rearrange to solve for R to yield R = 169-[(I^3-26)/(3I)]^3.  This must be greater than 0.

Then solving 169-[(I^3-26)/(3I)]^3 > 0 yields (I^3-104)*(I^3+13)^2 > 0, which means that I<cbrt(104)=4.70267.

Then integer values of I are limited to 1,2,3,4.  Corresponding values of R are then 20188/27, 196, 123200/729, and 29645/216.  These are all the non-negative values of R for which I is an integer.

 Posted by Brian Smith on 2016-07-09 11:27:56

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