Question

The prime
factorisation of P and q is not of the form 2

العنخ​

Answer 1

your question is not clear mate

Answer 2

Answer:

here's ur answer

Step-by-step explanation:

Here is a proof based on quadratic reciprocity.

Suppose p=3k−1|a2−ab+b2 with a,b relatively prime. Neither a nor b can have zero residue modp, else they both would which the hypothesis denies. Thus there exists a nonzero residue r≡ab−1modp such that r2−r+1≡0. Multiplying by 4 and completing the square gives (2r−1)2≡−3. Therefore the Legendre symbol (−3|p)=+1 if p is as assumed above.

But (p|3)=(3k−1|3)=−1 and −3≡+1mod4, so QR forces the contradictory conclusion (−3|p)=−1.

The case p=3 remains and this is a bit tricky. There are multiples of 3 having the form a2−ab+b2 where a and b are relatively prime. We just need a≡−bmod3 because a2−ab+b2≡(a+b)2mod3. But, if a≡−bmod3 with the common residue being 1 or 2, we find that c2≡3mod9 which doesn't work.

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