Let $p \equiv 1 \pmod 4$. Show that the Legendre symbols $(\frac{2a}{p}) = (\frac{a}{p})$ where $a$ is odd and $p \not \mid a$.
I believe this is a false statement. The statement is true for $p \equiv 1 \pmod 8$.
The following is a counter-example.
Counter-Example
We take $p=5$ which satisfies $p \equiv 1 \pmod 4$.
We also take $a=3$ which satisfies the condition $a$ is odd, and $p \not \mid a$.
The Legendre symbols are as follows.
$$ (\frac{2 \times 3}{5}) = (\frac{2}{5}) \times (\frac{3}{5}) = -1 \times (\frac{5}{3}) =-1 \times (\frac{2}{3}) = -1 \times -1 = 1 $$
$$ (\frac{3}{5}) = (\frac{5}{3}) = (\frac{2}{3}) = -1 $$
So $(\frac{2a}{p}) = (\frac{a}{p})$ does not hold under the given conditions.
Note: the author's official solution considers $p \equiv 1 \pmod 8 \implies p \equiv 1 \pmod 4$ but does not consider $p \equiv -3 \pmod 8 \implies p \equiv 1 \pmod 4$.