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Hi, I've been studying some proofs of The four squares theorem. Some of them are pretty clear. However, I came across a statement that The four squares theorem can be easily derived from Gauss-Legendre three squares theorem. Hard as I tried, I couldn't find out how to do it.

I was hopping someone could give me some idea or point out some article.

Thank you!

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5  
0 and 1 are both squares... –  zeb Mar 16 '13 at 10:31
    
jstor.org/stable/2033735 This gives a short and elementary proof of three square theorem. –  i707107 Mar 16 '13 at 16:54

2 Answers 2

The three squares theorem tells you that a positive integer $n$ can be represented as the sum of 3 squares if and only if it is not of the form $n = 4^a(8b+7)$, see e.g. http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares. Thus it suffices to show that for every integer of the form $n = 4^a(8b+7)$, there is an $m \in \mathbb{N}$ such that $n - m^2$ is not of this form. However taking $m := 2^a$, we get $n - m^2 = 4^a(8b+6)$, and we are done.

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The page you linked to just proves the easy part of the three squares theorem. –  Marc van Leeuwen Mar 16 '13 at 13:42
3  
@Marc: The OP asked for a proof of the implication "three squares theorem => four squares theorem" rather than for a proof of the three squares theorem. –  Stefan Kohl Mar 16 '13 at 14:06
    
That's really neat. Thank you:) –  user32252 Mar 17 '13 at 9:09

Also, the Gauß/Legendre theorem in question implies the following (improved) version of the four-squares theorem:

Every positive integer is the sum of four POSITIVE squares unless it belongs to the set

$A \cup B$

where

$A$ $=$ {$1, 3, 5, 9, 11, 17, 29, 41$}

and

$B$ $=$ {$2\cdot 4^{m}: m \in \mathbb{Z}^{+}$} $\cup$ {$6\cdot 4^{m}: m \in \mathbb{Z}^{+}$} $\cup$ {$14\cdot 4^{m}: m \in \mathbb{Z}^{+}$}.

Proof (d'après Prof. J. H. Conway in "The sensual quadratic form" [page 140]). By Gauß/Legendre and the well-known result on numbers that can be written as a sum of two squares it follows that every natural number of the form $8k+3$ (or $8k+6$) is the sum of three positive squares. Multiplying by $4$, we see that the same conclusion applies to natural numbers of the forms $32k+12$ and $32k+24$. Then, one can show that any integer $>49$ which is not a multiple of $8$ is the sum of four positive integers by subtracting an square so as to obtain a number of one of the aforementioned forms (for instance, from a number of the form $8k+2$ subtract $2^{2}$ in order to obtain a number of the form $8k+6$...). The proof is completed by checking the numbers up to $49$ and verifying that a number $n$ divisible by $8$ is the sum of four positive squares only if $\frac{n}{4}$ is. QED.

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