The Four Squares Theorem says that every natural number is the sum of four squares in $\mathbb Z$. What is known about coprime representations? Here we call a presentation $n=a^2+b^2+c^2+d^2$ coprime if the g.c.d. of the four numbers $a,b,c,d$ is 1. Does every natural number have a coprime presentation? If not, is there a simple criterion characterising the numbers that have coprime presentations? What is known about the number of different coprime presentations of a given $n$?

1$\begingroup$ If $8n$ then you must have all the four variables being even. Apart from that you can establish the result by Mobius inversion and using Jacobi's result on sums of four squares. $\endgroup$ – Lucia Nov 27 '17 at 6:42

2$\begingroup$ @Lucia: Could you please explain that? I don't quite get it, sorry. $\endgroup$ – user1688 Nov 27 '17 at 6:53

$\begingroup$ If $n$ is odd (for simplicity) and you write $n = rs^2$ where $r$ is squarefree, then your question is equivalent to checking whether $\sum_{a  s} \mu(a) \sum_{b  r(s/a)^2} b \geq 1$. The arithmetic seems annoying... but if you further assume that $\gcd(r,s) = 1$ then this is equivalent to checking whether $\sum_{a  s} \mu(a) \sigma((s/a)^2) \geq 1$, since $\sigma(r)$ would factor out (here $\sigma$ is the sum of divisors function). Note that $\sigma(s^2)$ is also multiplicative, so Mobius inversion would work $\endgroup$ – Stanley Yao Xiao Nov 27 '17 at 9:50
Let $R(n)$ denote the number of ways of writing $n$ as a sum of $4$ squares, and $r(n)$ the number of ways where gcd of $(a,b,c,d) =1$. Then grouping representations of $n$ as a sum of $4$ squares according to the gcd of the variables, clearly we have $$ R(n) = \sum_{k^2  n} r(n/k^2), $$ and so by Mobius inversion $$ r(n) = \sum_{k^2 n} \mu(k) R(n/k^2). $$
Now by Jacobi's four square theorem, $R(n)$ is given explicitly as $8$ times a multiplicative function $F(n)$ defined on prime powers by
$$
F(2^k) = 3 \text{ for all } k\ge 1,
$$
and, for odd primes $p$,
$$
F(p^k) = p^k + p^{k1} + \ldots + 1.
$$
So $r(n)$ is $8$ times a multiplicative function $f(n)$ which is defined on prime powers by
$$
f(2)= 3; \ \ f(4) = 2;\ \ f(2^k)=0 \text{ for } k \ge 3,
$$
and for odd primes $p$ and $k\ge 1$
$$
f(p^k) = p^k + p^{k1}.
$$

1$\begingroup$ Good to have this on record! For the interested reader: there is a similar (but somewhat deeper) formula for the sum of 3 coprime squares. See our responses at mathoverflow.net/questions/217698/… $\endgroup$ – GH from MO Nov 28 '17 at 17:52

$\begingroup$ I am wondering if there is a formula for the number of representations where every summand is coprime to any other. $\endgroup$ – Dr. Pi Dec 22 '17 at 15:42
As Lucia noticed, if $8\mid n$, then $a,b,c$, and $d$ must be even, so that there is no "coprime representation". If $8\nmid n$ with $n$ large enough, one can argue as follows.
Find a positive integer $d<\sqrt n$ such that
 $d$ is coprime with $n$;
 $d$ and $n$ are of different parity;
 moreover, if $n\equiv 3\pmod 8$, then $d\equiv 0\pmod 4$, and if $n\equiv 7\pmod 8$, then $d\equiv 2\pmod 4$.
One readily verifies that $d^2\not\equiv n+1\pmod 8$, as a result of which $nd^2$ is not of the form $4^u(8v+7)$ (where $u$ and $v$ are nonnegative integers). Therefore, by Legendre's threesquare theorem, $nd^2$ is a sum of three squares: $nd^2=a^2+b^2+c^2$. This yields $n=a^2+b^2+c^2+d^2$, and coprimality of $d$ and $n$ ensures that $\gcd(a,b,c,d)=1$.