Positive integers $4^{a}m$ with $m\equiv 7(\mathrm{m}\mathrm{o}\mathrm{d}8)$ are not sums of three integer squares

Let $a\in \mathbb{N}$ and let $m$ be a positive integer with $m\equiv 7(\mathrm{m}\mathrm{o}\mathrm{d}8)$ (Congruence modulo an integer: $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}n)$ when $n\mid (a-b)$, including the moduli $0$ and $1$). Then there are no integers $x,y,z$ with $4^{a}m=x^2+y^2+z^2$, where $4^a$ is the natural power of $4$ in the commutative monoid $(\mathbb{Z},\cdot ,1)$ (Powers $g^n$: natural exponents in a monoid and integer exponents in a group, with $g^0=e$, $(\mathbb{Z},\cdot ,1)$ is a commutative monoid whose group of units is $\{1,-1\}$; equivalently $u\mid 1$ holds exactly for $u=1$ and $u=-1$).

Three descriptions of the same integers. For a positive $m$, the condition $m\equiv 7(\mathrm{m}\mathrm{o}\mathrm{d}8)$ says $m=8b+7$ for an integer $b$, and $b\ge 0$ because $8b+7>0$; so the integers excluded here are exactly those of the form $4^a(8b+7)$ with $a$ a natural number and $b$ a nonnegative integer, which is how Dummit writes them. Crisman's phrase for the same set, an even power of two times an odd number congruent to seven modulo eight, is a third description: $4^a=2^{2a}$ and every $m\equiv 7(\mathrm{m}\mathrm{o}\mathrm{d}8)$ is odd.

Only one direction is proved. The statement says these integers are not sums of three squares. Its converse, that every other nonnegative integer is a sum of three squares, is Legendre's three-square theorem; it is not available from this page's declared prerequisites, and nothing here uses it. In particular the argument above rules out no integer beyond the ones named.

Why the induction is on the exponent. The base case is a congruence computation modulo $8$ and nothing more. The step is where the work is: it needs that a sum of three squares divisible by $4$ has all coordinates even, which is If $4$ divides $x^2+y^2+z^2$ then $x$, $y$ and $z$ are all even, since without it the halved coordinates need not be integers.