Lucky Lattice Points

Firstly, note that $2013\equiv 13\pmod{16}$ and that the only squares modulo $16$ are $0,1,4,9$ . The only way to get $13$ from these is $0+4+9$ . Suppose for now that $x$ is the odd number, i.e. that $x^2\equiv 9\pmod{16}$ . Analysing which $x$ satisfy this congruence, we deduce that $x\equiv 3,5\pmod{8}$ . Before checking these $x$ 's case by case, we eliminate a few more of them.

Note that $2013=3\times 11\times 61$ . In particular $x^2+y^2+z^2$ is divisible by $3$ and $11$ , but not by their squares. I claim that this implies that none of $x,y,z$ can be divisible by either $3$ or $11$ , since both of these leave a remainder of $3$ when divided by $4$ .

To prove the claim, let $p\equiv 3\pmod{4}$ be a prime, suppose that $p$ divides $x^2+y^2+z^2$ , that $p^2$ does note divide it, and that $p$ divides $x$ . This means that $p$ divides $y^2+z^2$ , but that $p^2$ does not. However, since $p\equiv 3\pmod{4}$ , if $p$ divides $y^2+z^2$ , then $p$ divides both $y$ and $z$ , contradicting the fact that $p^2$ does not divide $y^2+z^2$ .

We now check case by case the odd numbers $x$ from $1$ to $43$ (which is the largest odd number less than $\sqrt{2013}$ ) which are $\equiv 3,5\pmod{8}$ and are not divisible by $3$ or $11$ . This leaves $x\in\{5,13,19,29,35,37,43\}$ . For each of these $2013-x^2=y^2+z^2$ is the sum of two even squares, and hence $\frac{2013-x^2}{4}$ is also the sum of two squares. For those values of $x$ , the corresponding values of $\frac{2013-x^2}{4}]$ are $\{497,461,413,293,197,161,41\}$ . Among these we have $497=7\times 71, 413=7\times 59, 161=7\times 23$ , while the other $4$ are prime. We have seen that if $7$ divides $y^2+z^2$ , then $7^2$ divides it, ruling out the first three options. On the other hand, the four primes are congruent to $1$ modulo $4$ , meaning that each is uniquely expressible as the sum of two squares (not counting permutation).

Hence there are $4$ solutions where $(x,y,z$ have a certain ordering and are all positive. They can be rearranged in $3!=6$ ways, and minus signs can be introduced in $2^3=8$ ways. This gives a total of $4\times 6\times 8=192$ triples.

We will first assume that $x, y, z$ are positive, and not care about the order.

Working modulo 4, since $2013 \equiv 1 \pmod {4}$ , and the quadratic residues modulo 4 are 0 and 1, it follows that exactly 1 of the integers is odd. Working modulo 16, $2013 \equiv 13 \pmod{16}$ , and the quadratic residues modulo 16 are 0, 1, 4, 9, we can only get $13 = 0 + 4 + 9$ . Hence, the numbers must be of the form $4A, 8B \pm 2, 8 C \pm 3$ .

We know that the largest possible integer is $\sqrt{2013} < 45$ . Let's check for possible values of $C = 3, 5, 11, 13, 19, 21, 27, 29, 35, 37, 43$ . Recall Fermat's 2 square theorem, which tells us that a number can be written in the form $a^2 + b^2$ , if and only if every prime of the form $4k+3$ appears an even number of times in the expansion.

$2013 - 3^2 = 2004 = 4 \times 3 \times 167$ will not work.
$2013 - 5^2 = 4 \times 7 \times 71$ will not work.
$2013 - 11^2 = 4 \times 11 \times 43$ will not work.
$2013 - 13^2 = 4 \times 461$ yields the solution $(38, 20, 13)$ .
$2013 - 19^2 = 4 \times 7 \times 59$ will not work.
$2013 - 21^2 = 4 \times 3 \times 131$ will not work.
$2013 - 27^2 = 4 \times 3 \times 107$ will not work.
$2013 - 29^2 = 4 \times 293$ yields the solution $(34, 4, 29)$ .
$2013 - 35^2 = 4 \times 197$ yields the solution $(28, 2, 35)$ .
$2013 - 37^2 = 4 \times 7 \times 23$ will not work.
$2013 - 43^2 = 4 \times 41$ yields the solution $(10, 8, 43)$ .

Hence, there are 4 triples of positive numbers that work. Since the numbers are distinct, and we care about the order, there are $4 \times 6 = 24$ triples of positive numbers. Since each sign could be positive or negative, there are $24 \times 8 = 192$ lattice points on the sphere.

For easier manipulation we first consider a,b,c to be all non-negative integers.

First we note that squares of even integers give 0 (mod 4) and squares of odd integers give 1 (mod 4). 2013 is congruent to 1 mod 4, so we know that only one of a,b,c is even. Without loss of generalization, let a=2x, b=2y and c=2z+1.

We can rewrite the equation as (2x)^2 + (2y)^2 + (2z+1)^2 = 4x^2 + 4y^2 + 4z(z+1) + 1 = 2013. Subtracting 1 from both sides and dividing them by 4, we get x^2 + y^2 + z(z+1) = 503.

Now note that z(z+1) can only be either 0 mod 4 (when z is 0 or 3 mod 4) or 2 mod 4 (when z is 1 or 2 mod 4). Since 503 is 1 mod 4, the only possibility is that x and y are both odd, and z is 1 or 2 mod 4. We thus split into 2 cases.

If z is 1 mod 4, we let x=2m, y=2n and z=4k+1. Expanding the equation we get 4m^2 + 4n^2 + 4n + 1 + 16k^2 + 12k + 2 = 503, and after subtracting 3 on both sides and dividing by 4 we get m^2 + n(n+1) + k(4k+3) = 125.

If z is 2 mod 4, we let x=2m, y=2n and z=4k+2. Expanding the equation we get 4m^2 + 4n^2 + 4n + 1 + 16k^2 + 20k + 6 = 503, and after subtracting 7 on both sides and dividing by 4 we get m^2 + n(n+1) + k(4k+5) = 124.

So we get m^2 + n(n+1) = 125 - k(4k+3) or 124 - k(4k+5).

Since a,b,c are non-negative integers, m,n,k are non-negative integers as well. Therefore, k = 0 to 5 for the first case and 0 to 4 for the second. We split into cases where k = 0 to 5, where we can easily check for solutions.

When k = 5,

m^2 + n(n+1) = 10. This gives us a possible solution, when m=n=2.

When k = 4,

m^2 + n(n+1) = 40 or 49. This gives us a possible solution, when m=7, n=1.

When k = 3,

m^2 + n(n+1) = 73 or 80. This gives us a possible solution, when m=1, n =8.

When k = 2,

m^2 + n(n+1) = 98 or 103. There is no nonnegative integers m,n that satisfy the equation.

When k = 1,

m^2 + n(n+1) = 115 or 118. This gives us a possible solution, when m=5, n = 9.

When k = 0,

m^2 + n(n+1) = 124 or 125. There is no nonnegative integers m,n that satisfy the equation.

This gives us a total of 4 solutions for (m,n,k), where m,n,k are all positive. However, a,b,c are interchangeable, which gives us 4x6=24 solutions for (a,b,c) where all of a,b,c are positive.

Now note that any possible solution (p,q,r) would mean that the (± p, ± q, ± r) is a solution, and vice versa, so we have a total of 24x8=192 solutions in all.

Use stars and bars method to find positive solutions. And since it's all variables squared, find permutations of negative solutions as well.