A density one problem for diophantine equations

I read that the set of integers that can be written in the form $$n = a^2 + b^4 + c^6$$ is of zero density, since the sum of inverses of exponents $1/2+1/4+1/6$ is less than $1$. I do not understand the argument, is there a probabilistic/density way of seeing/writing this?

Solution 1:

I don't know much about any theory behind this, but to me it looks like simple counting.

The number of squares less or equal than $n$ is $O(n^{1/2})$. (In fact, it is easily calculated as $\lfloor\sqrt{n}\rfloor$.) Similarly, the number of fourth powers in the same interval is $O(n^{1/4})$ and the number of sixth powers in the same interval are $O(n^{1/6})$. Now, make up all the sums of any square, any fourth power and any sixth power in that set. The number of those sums is at most $O(n^{1/2})O(n^{1/4})O(n^{1/6})=O(n^{1/2+1/4+1/6})$ (there may be repetitions!) and so the number of those sums less than equal than $n$ (which is bound from above by the previous number, and may be even smaller if some of the sums end up bigger than $n$) is also $O(n^{1/2+1/4+1/6})=o(n^1)=o(n)$, because $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}<1$.

As the density is defined as the limit of the fraction of $\{1,2,\ldots, n\}$ belonging to the set, when $n\to\infty$, we have that the density here is $\lim_{n\to\infty}\frac{o(n)}{n}=0$.