Show that $a \equiv \pm 3\pmod 8$ is $a^2 \equiv 9 \pmod {16}$.

Question

Show that $a \equiv \pm 3 \pmod 8$ is $a^2 \equiv 9 \pmod {16}$.

My attempt is incomplete:
$a \equiv \pm 3 \pmod 8 \implies (a \equiv 3 \pmod 8) \cup (a \equiv - 3 \pmod 8) $
It is possible to take only one value, not both if the modulus is the same, for a given value $a$. The reason for the given example, considering both as possible would lead to :
$3 \pmod 8 \equiv -3 \pmod 8 $
$\implies 3 + 8n = 5 +8m, \exists n,m \in \mathbb{Z}$
$\implies 2 = 8(n-m)$
$ \implies 1 = 4 (n-m) $
$\implies (n-m) = 1/4$
which is impossible for both $n,m$ being integers.

The difficulty is in finding the logic behind using the two possible (but, mutually exclusive) solutions.

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Based on answers below, have identified the given solution approach that concerns with finding the original equation that can have either of the two roots. This original equation should have both roots as possibility. Hence, the answers below emphasize multiplication of the two sub-equations of the original equation, & grouping together the terms with common factor as $16$ and residue $9$.

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I am not fully sure of the below issue being clear to me, and am restating it:

But have a hitch in understanding given answer(s), as I compare the original equation's derivation, based on the two mutually exclusive solutions, to the derivation of a quadratic equation from two roots. But in this question, the two roots are mutually exclusive. Does it not make multiplication un-obvious.
Is it that the quadratic equation has roots, while here each residue class is its own equation. This is why for a given modulo $m$, there are stated to be $m$ in-congruent solutions, that form a complete set of residues modulo $m$.
Hence, no parallel can be drawn between the two: - quadratic equation and two different residue values for a modular equation.

Answer 1

Your statement $$a \equiv \pm 3 \pmod 8 \implies (a \equiv 3 \pmod 8) \wedge (a \equiv - 3 \pmod 8)$$ is not logically true. You may change it to $$a \equiv \pm 3 \pmod 8 \implies (a \equiv 3 \pmod 8) \lor (a \equiv - 3 \pmod 8)$$

Note that $$(a=8k\pm 3)\implies ( a^2 = 64 k^2 +9\pm 48k)=16M+9$$Thus the statement is true.

Answer 2

If $a=8k\pm3$ then $a^2=64k^2\pm48k+9=16(4k^2\pm3k)+9$ how hard is this?