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.
Solution
