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Lagrange showed that every natural number can be represented as the sum of 4 squares (of natural numbers). Let us call a natural number a Descartes number if it can be represented as the sum of the squares of 4 natural numbers, with the natural numbers being the curvatures of 4 circles in Descartes configuration (that is, that fit the hypothesis of the Descartes Circle Theorem).

That is, call a natural number $n$ a Descartes number if there exist natural numbers $k_1,k_2,k_3$ and $k_4$ such that $$ n = k_1^2+k_2^2+k_3^2+k_4^2=\frac{1}{2}(k_1+k_2+k_3+k_4)^2.$$

Do Descartes numbers exist? More generally, what is their density?

Here is a related link:

http://www.e-bookspdf.org/view/aHR0cDovL3d3dy50aGVvcmVtb2Z0aGVkYXkub3JnL0dlb21ldHJ5QW5kVHJpZ29ub21ldHJ5L0Rlc2NhcnRlc0NpcmNsZS9Ub3RERGVzY2FydGVzQ2lyY2xlLnBkZg==/VGhlb3JlbSBPZiBUaGUgRGF5

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  • $\begingroup$ Doesn't the link you give show that 4050 is one such number (so they exist)? $\endgroup$ – Matthew Conroy Dec 2 '13 at 21:05
  • $\begingroup$ I added what I think is a more explicit definition of your "Descartes number". Please check it to make sure this is what you mean. Thanks. $\endgroup$ – Matthew Conroy Dec 2 '13 at 22:20
  • $\begingroup$ The first thing I'd want to have, to attack your question, is a parametric formula for all integer quadruples $(k_1,k_2,k_3,k_4)$ that satisfy that equation (similar to a parametric formula for all Pythagorean triples of integers). $\endgroup$ – Greg Martin Dec 2 '13 at 22:55

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