Consider this paper written by Lubotzky, Phillips and Sarnak (1986) on expander graphs.
Below definition 2.2, they let $p,q$ be primes both congruent to $1\mod{4}$. Then they claim that there are $p+1$ vectors $a=(a_0,a_1,a_2,a_3)$ where $a_0\in\mathbb{N}$, $a_j\in\mathbb{Z}$, $a_0$ is odd and $a_j$ even for $j=1,2,3$, and $a_0^2+a_1^2+a_2^2+a_3^2=p$.
Regretfully I have learned little about number theory and I can't figure out why there exists (maybe exactly?) $p+1$ such vectors.
Hope I have clearly expressed my question and thanks for your help!
Jacobi's four square theorem says that the number of vectors $(w,x,y,z)$ for which $w^2+x^2+y^2+z^2=p$ is exactly $8(p+1)$, where there are no restrictions on $w,x,y,z$. When $p$ is 1 modulo 4, exactly one of the entries is odd, and since it is nonzero, is either positive or negative; say it is $x$, so that $(w,x,y,z)$ and $(w,-x,y,z)$ are both valid vectors. There are four positions in which to place the odd entry, and there are two possibilities for the sign of that entry, so if we want our vector to have the same shape as LPS requires (positive odd first followed by even) the Jacobi guarantee overcounts by a factor of 8.