p is prime and $p=4t+3$, I am wondering how to prove

Question

1. when $p ≡ 1\pmod7$ $$(-1)^\frac{(p+1)}{4} = 1$$

2. when $p ≡ 5 \pmod3$ $$(-1)^\frac{(p+1)}{4} = -1$$

My attempt: For question 1, $p=4t+3$,hence $$\frac{(p+1)}{4} = t+1$$ $4t+3≡1\pmod7$, hence $$t+1≡4\pmod7$$

For question 2, $p=4t+3$,hence $$\frac{(p+1)}{4} = t+1$$ $4t+3≡5\pmod3$, hence $$t+1≡0\pmod3$$ Then I do not know how to prove it at all. What do you think about it? Could you please show me?

Regards

Answer 1

Both statements are false.

For example, $p=43$ satisfies $p=4t+3$ and $p=1\pmod{7}$ but $(-1)^{11}=-1$.

Similarly, $p=23$ satisfies $p=4t+3$ and $p=2\pmod{3}$ but $(-1)^{\frac{p+1}{4}}=(-1)^6=+1$.