Sums of three non-zero squares - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T18:42:52Zhttp://mathoverflow.net/feeds/question/90914http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/90914/sums-of-three-non-zero-squaresSums of three non-zero squaresAndres Caicedo2012-03-11T17:02:02Z2012-03-11T22:15:26Z
<p>It is a well-known result of Legendre that a positive integer is sum of three squares unless it is of the form $4^a(8b+7)$.</p>
<p>In </p>
<ul>
<li>Grosswald, E.; Calloway, A.; Calloway, J. <em>The representation of integers by three positive squares</em>. Proc. Amer. Math. Soc. 10 1959 451–455. (MR0104623 (21 #3376)),</li>
</ul>
<p>it is shown that there is a finite set $T$ such that any positive integer is a sum of three <em>non-zero</em> squares unless $n$ is of the form $4^a(8b+7)$ or of the form $4^am$ where $m\in T$.</p>
<p>The set $T$ is essentially identified, see </p>
<ul>
<li>Grosswald, Emil. <strong>Representations of integers as sums of squares</strong>. Springer-Verlag, New York, 1985. xi+251 pp. ISBN: 0-387-96126-7 (MR0803155 (87g:11002)):</li>
</ul>
<p>Either
<code>$$ T=\{1,2,5,10,13,25,37,58,85,130\}, $$</code>
or else the Riemann hypothesis fails, and $T$ consists of these 10 numbers, and <em>at most</em> another one, $k$, that must be larger than $5\cdot10^{10}$. The conjecture is that $|T|=10$, of course. </p>
<p>I could not find any updates on the question of whether the conjecture has been settled, and would appreciate any information or pointers to the relevant literature.</p>
http://mathoverflow.net/questions/90914/sums-of-three-non-zero-squares/90941#90941Answer by Barry Cipra for Sums of three non-zero squaresBarry Cipra2012-03-11T22:15:26Z2012-03-11T22:15:26Z<p>Googling on the title of the Grosswald paper produced a link to math.uab.edu/~simanyi/Goswick_et_al_final.pdf which (backing up the url to the ~simanyi) indicates it's a recent paper in JNT. (Re-posted from comments at the OP's suggestion.)</p>