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Lagrange's four-square theorem states that every natural number can be represented as the sum of four integer squares.

$$p=a_{0}^{2}+a_{1}^{2}+a_{2}^{2}+a_{3}^{2}$$ where the four numbers $a_{0},a_{1},a_{2},a_{3}$ are integers (source).

I would like to calculate as few square numbers as possible.

Jacobi's theorem: there are multiple combinations of $[a_{0},a_{1},a_{2},a_{3}]$ for a number (source).

Eduard Wirsing proved that there exists a set of squares S with: $$|S|=O(n^{1/4}\log^{1/4}n)$$ such that every positive integer smaller than or equal n can be written as a sum of at most 4 elements of S (source). Note: I don't have access to this article.

Is it $(\log n)^{1/4}$? Is it $\log_{2}$ or $\ln$ ? Note: it's not $\log_{1/4}(n)$ gives negative numbers. I suppose that the formula is: $${|S|=O(n^{1/4}(\ln n)^{1/4})}$$

Could you confirm me that, by example, for: $n=2^{10}=1024$ $$O(1024^{1/4} \ln(1024)^{1/4})=9,18 => 10$$ Every number between $0$ and $1024$ can be represented by the square numbers between: $0$ and $10$.

$$2^{10} = 1024 = 10^2 + 4^2 + 2^2 + 2^2$$

Is there a better result than $|S|=O(n^{1/4}(\ln n)^{1/4})$?

PeterMacGonagan
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    About the $\log$s, the base of the logarithm will not affect the O estimate. – orangeskid Jul 09 '22 at 11:34
  • This is not clear. Do you want to bound the largest used perfect square ? – Peter Jul 09 '22 at 12:01
  • $O(1024^{1/4} \ln(1024)^{1/4})=9$ is wrong. That's not how $O$ bounds works. Saying that $|S| = O(n^{1/4}(\log n)^{1/4})$ means there's exists numbers $M$ and $N$ such that $|S| \le M (n^{1/4}(\log n)^{1/4})$ for all $n \ge N$. We we often don't know the explicit values of $M$ and $N$. So evaluating at $n=2^10$ doesn't work because it could be that $N = 2^{1000000}$ so the bound only works for $n$ bigger than that. Or it could be that $N = 1$, and in that case the bound do works for $n=2^10$, but may be $M = 10000$ so the bound becomes trivial and useless for that case. – jjagmath Jul 09 '22 at 12:12
  • @Peter yes, I would like to bound the largest used perfect square. – PeterMacGonagan Jul 09 '22 at 12:38
  • @jjagmath I wasn't aware to that... How to know the largest perfect square to use? Is there a method? – PeterMacGonagan Jul 09 '22 at 12:40

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