A prime $p = 2a^2 – 2ab + 3b^2$ for integers $a$ and $b$, then $p$ cannot be congruent to $11 \pmod{20}$. 
A prime $p = 2a^2 – 2ab + 3b^2$ for integers $a$ and $b$, then $p$ cannot be congruent to $11 \pmod{20}$.

I need to prove the conclusion and I tried by constructing a contradiction:
First, I suppose $$p \equiv 11 \pmod{20}$$ and factorized the prime as $$(a+3b)(a-b)$$
So I get two possible cases:

*

*$(a+3b)\equiv 1 \pmod{20}$ and $(a-b) \equiv 11 \pmod{20}$


*$(a+3b)\equiv 11\pmod{20}$ and $(a-b) \equiv 1 \pmod{20}$
However I do not know how to deal with it.
Is this a right direction or there are other good methods to prove the statement? Looking forward to any suggestions!

We have clearly that $2|b-1$ so it is clear aswell that $4|2a(a-b)$ so we have
$$4|p-3$$
in order to prove the problem , we just need to show the following doesnt hold
$$5|p-1$$
now for the sake of contradiction , we assume it is true
which gives
$$5|2p-2=(2a-b)^2+5b^2-2$$
so we need $2$ to be a square in $Z_5$ which is wrong.
so it is impossible to have
$$20|p-11$$
