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!
 A: Well, initially I didn't see that your factorization was wrong, so I have written the answer according to the cases you got in the end. This is how you would do If the factorization was correct.
Just subtract the two equations you got to get $$4b \equiv \pm 10 \pmod{20}$$ which means $4b\pm 10$ is divisible by $20$, but this is not possible since $20$ is of the form $4k$.
A: 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
$$20|p-11$$
A: If $p$ is congruent to 11 (mod 20) then it is congruent to 1 (mod 5). Then the result follows from $2p = (2a-b)^2 + 5b^2$, which implies that $(2a-b)^2 \equiv 2 \mod 5$, which is impossible.
A: The only possible values of $(2a^2 – 2ab + 3b^2) \bmod 20$ are $0,2,3,7,8,10,12,15,18$.
