More fun in 2015, Part 3

The point $21,5,2015$ is trivial and can be assumed to be the origin. What we need to find is the number of integer solutions for:

$x^2 +y^2 +z^2=2015$

Observe that a square of an integer is always either $0,1$ or $4$ modulo $8$ . Therefore, the sum of three squares can be $(0,1,2,3,4,5)$ or $6$ modulo $8$ . You can check this out yourself, no combination of three remainders will ever give you a number that is $7$ modulo $8$ .

$2015= 251\times 8+7$

I rest my case.

Similar to the other solutions, we can disregard the point $(21,5,2015)$ completely. We are only interested in $x^2+y^2+z^2=2015$ .

Consider the equation $x^2+y^2+z^2=2015$ . Note that $a^2=0,1(mod 4)$ , but $2015 = 3 (mod 4)$ . Thus, all of x, y, and z must be odd. Replacing $(x,y,z)$ with $(2a+1,2b+1,2c+1)$ respectively, we have the equation $(2a+1)^2+(2b+1)^2+(2c+1)^2 = 2015$ , and expanding and simplifying gives $a(a+1)+b(b+1)+c(c+1)=503$ . But we know that exactly one of $(x,x+1)$ must be even, so the left side of the above equation must be even for any integer $x$ , but the right side of the above equation is odd. Hence, there are no possible solutions of $(a,b,c)$ , and hence no possible solutions for $(x,y,z)$ .

Legendre's three-square theorem states that

A natural number $n$ can be only expresses as a sum of three squares of integers ( $n = x^2 + y^2 + z^2$ ) if and ony if

$n$ is not equal to $4^a(8b + 7)$ where $a$ and $b$ are some integers