Note that $n \in Z$ if it satisfies the equation. The function $n+ \lfloor \sqrt{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor$ is strictly increasing in integers. Therefore there can be only one value for which the value of the function is 2014. That value is 1958 which one can check by trying values. Hence that is the only answer.

Other solution?

VicKmath7 wrote: Other solution? Are you really asking for a longer and more complex solution ?

No, just not with functions

Well, you didn't explain how you check values. You can use binary_search(). Here is the explanation of binary_search(). https://en.wikipedia.org/wiki/Binary_search_algorithm

It's probably faster to just start at $2014$ and keep subtracting off how much higher the result we get is than it should be.

I think it's quite natural to try $2000$ and to observe that the result is not much bigger than $2014.$ Or it might be enough to just note that $n<2014.$ After that, we can try to find an interval of the form $[k^2,(k+1)^2)$ where $n$ lies and based on the above observation, we might want $(k+1)^2=2025,$ so $k=44,$ which works because for $n=44^2$ the given expression is smaller than $2014.$ Now, for $n \in [44^2,45^2)$ we know that $\lfloor \sqrt{n} \rfloor=44$ and $\lfloor \sqrt[3]{n} \rfloor=12,$ so the given expression is $n+56$ for all $n \in [44^2,45^2)$ From $n+56=2014$ we get $n=1958,$ which is the only one solution since the given expression is strictly increasing in $n.$

First, we try to make x^6<n<(x+1)^6. We get that 3^6<n<4^6 So we get Floor[Sqrt[44^2]] + Floor[CubeRoot[12^3]] equals to n+44+12=2014 -->n=1958

Apex_master wrote: First, we try to make x^6<n<(x+1)^6. We get that 3^6<n<4^6 So we get Floor[Sqrt[44^2]] + Floor[CubeRoot[12^3]] equals to n+44+12=2014 -->n=1958 Sorry, im not so good at explaining mathmatic solution in english

As $n<2014$ so $[\sqrt{n}] +[\sqrt[3]{n}]\leq 56$ and observe for all $1936\leq n \leq 2014$ $[\sqrt{n}] +[\sqrt[3]{n}]=56$ hence we get a solution at $n=2014-56=1958$ and after this point as $[\sqrt{n}] +[\sqrt[3]{n}]<56$ and $n<1936$ hence no more solution is possible!!

$f(n)=n+ \lfloor \sqrt{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor$ is strictly increasing over $\mathbb Z$, so there is a maximum of one solution. Testing (using the strictly increasing condition) we find that the only solution is $\boxed{1958}$.

parmenides51 wrote: Solve in positive real numbers: $n+ \lfloor \sqrt{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor=2014$ Who is the author?

wizixez wrote: parmenides51 wrote: Solve in positive real numbers: $n+ \lfloor \sqrt{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor=2014$ Who is the author? Zlatko Salko also known as the SHEF Here is the link: https://www.youtube.com/watch?v=pBfh0DpK2yg&list=PLnwFx3rpTMsvi_uep0RucW0NGxLVgu3Nr&index=3

Thanks..