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This is most probably asked here already in some other form (as it seems to be a very basic question) but since I have't found some question very similar to this one I feel obliged to ask (also because I do not know how to answer it at present and because this question appears in some of elementary researches I do).

Lagrange's four-square theorem guarantees that for every $m \in \mathbb N_0=\{0,1,2,...\}$ there is at least one quadruple of integers $(a_1(m),a_2(m),a_3(m),a_4(m))$ such that $m=\displaystyle\sum_{i=1}^{4}(a_i(m))^2$.

But, if the statement is slightly changed so to only allow quadruples that are subsets of $\mathbb N^4$ (instead of $\mathbb Z^4$ (or, equivalently, $\mathbb {N}_0^4$) then the statement is no longer true.

In other words, not every natural number is a sum of exactly four squares of naturals.

So, the set of all natural numbers (that is, elements of $\mathbb N=\{1,2,3,\dotsc\}$) that can be written as the sum of exactly four squares of natural numbers is not equal to all of $\mathbb N$.

If that exceptional set (the set of all natural numbers that cannot be represented as the sum of exactly four squares of naturals) is denoted as $E_4$, I have a basic question:

  • How to characterize elements of $E_4$?

For example, is there a statement of the form "$w \in E_4$ if and only if $w$ is (or is not) of the form $a^k(bl+c)$", for some special choice (or choices) of $a$, $b$, $c$?

Or there is some other characterization?

Edit: If I did not misinterpret something, it seems that we have an answer here, so the question can be appropriately closed.

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    $\begingroup$ I'm voting to close this question as off-topic because it was already answered on math.stackexchange. $\endgroup$ – Jeremy Rouse Apr 1 at 20:34
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    $\begingroup$ @JeremyRouse that seems appropriate, i have linked the MSE question, and the answer is there $\endgroup$ – user153451 Apr 1 at 20:35
  • $\begingroup$ If that´s better, I can also delete this question, upon request. $\endgroup$ – user153451 Apr 1 at 20:37
  • $\begingroup$ Sorry, I did not see Jeremy Rouse's remark. So I answered the question below. $\endgroup$ – GH from MO Apr 1 at 20:38
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    $\begingroup$ @GHfromMO That´s not a problem, it is good to have an answer, no matter the destiny of the question. I won´t delete it. $\endgroup$ – user153451 Apr 1 at 20:38
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This problem was solved by Dubois (1911). The exceptional integers are: $1$, $3$, $5$, $8$, $9$, $11$, $17$, $29$, $41$, $2\cdot 4^m$, $6\cdot 4^m$, $14\cdot 4^m$. This is quoted in this 2016 paper by Byeong Moon Kim.

I have not seen the original proof, but here is a sketch how I would prove it. First I would solve the problem for odd integers. An odd integer $n$ has at least $8(n+1)$ representations as a sum of four squares, but only $O(n^{1/2+\epsilon})$ representations as a sum of three squares (the implied constant depends effectively on $\epsilon>0$). Hence every sufficiently large odd integer is a sum of four positive squares.

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  • $\begingroup$ Is it foolish of me to guess that the every member of the exceptional set is a sum of either exactly two or exactly three squares of naturals (except 1)? $\endgroup$ – user153451 Apr 1 at 20:45
  • $\begingroup$ @Ante: $29=5^2+2^2=4^2+3^2+2^2$. Also, $1$ cannot be written as a sum of two or three positive squares. $\endgroup$ – GH from MO Apr 1 at 20:50
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    $\begingroup$ This is done quickly in Conway's little book, The Sensual Quadratic Form $\endgroup$ – Will Jagy Apr 3 at 1:55