8k + 7 is not a sum of three squares

Squares modulo 8. For $x$ odd, $x^2\equiv1\pmod8$; for $x\equiv2,6\pmod8$, $x^2\equiv4$; for $x\equiv0,4\pmod8$, $x^2\equiv0$. So squares are $0,1,4$ modulo $8$.

Sums of three. The sums of three elements of $\{0,1,4\}$ are:
$$0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 8\equiv0,\ 9\equiv1,\ 12\equiv4.$$
Precisely, the possible sums are the residues $0,1,2,3,4,5,6$ modulo $8$. The residue $7$ is never obtained (for instance $4+1+1=6$, $4+4+1=9\equiv1$, $1+1+1=3$).

Conclusion. No sum of three squares is $\equiv7\pmod8$.

Remark. Legendre's three-square theorem states the converse: a positive integer is a sum of three squares iff it is not of the form $4^a(8b+7)$. For example $15$, $23$, $28=4\cdot7$ are not sums of three squares.

Lesson. List the residues of squares modulo $8$ ($0,1,4$) and check which sums are attainable: the standard way to rule out representations.