# 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?

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 $$\lim_{n\to\infty}\frac{o(n)}{n}=0$$.
• it is also the case that $n=a^2 + b^2$ has density zero. Sad but true. Dec 6 '21 at 21:20
• @WillJagy A more refined argument can deduce that number of integers $\le N$ that can be expressed as sum of squares is asymptotic to $N(\log N)^{-1/2}$ multiplied by an explicit constant. Dec 7 '21 at 2:27