Question

Question No. 25 Let p and q be prime numbers such that p2 + 3pq+q2 = 5* for some positive integer k. Find the value of p? +2+k? please solve it fast!!!!!​

Answer 1

Step-by-step explanation:

For every odd prime number p; and for every positive integer α;

the multiplicative group Z∗pα;

is a cyclic group of order ϕ(pα)=(p−1)pα−1.

In other words:

(Z∗pα ,×)≡(Z(p−1)pα−1 ,+).

For p=2; and for every positive integer 3≤α;

the multiplicative group Z∗2α;

is the direct sum of Z2 and a cyclic group of order 12ϕ(2α)=2α−2.

In other words:

(Z∗2α ,×)≡(Z2⊕Z2α−2 ,+).

The multiplicative group Z∗22; is a cyclic group of order 2.

The multiplicative group Z∗2; is the trivial group.

If n is coprime with 3pq then we can factor n and hence we have:

n3pq≡3pqn⟹n3pq−1≡3pq1.

First case: All of 3,p,q are different. By the above lemma, we know that:

there is an integer a; with ordp(a)=p−1.

On the otherhand a3pq−1≡p1;

which implies that p−1∣3pq−1.

Similarly one can prove that p−1∣3pq−1 and 3−1∣3pq−1.

But notice that 3pq−1=3(p−1)q+3q−1;

similarly we have: 3pq−1=3p(q−1)+3p−1

and 3pq−1=2pq+pq−1.

So we can conclude that:

p−1∣3q−1 and q−1∣3p−1 and 3−1∣pq−1;

the last divisibility condition implies that both of p,q must be odd;

you can check that p=11,q=17 satisfies the above divisiblity conditions.

Answer 2

Step-by-step explanation:

Let p and q be prime numbers

and $p^2 +3pq +q^2 = 5$ (given )

Using AM ≥ GM

$\frac{p^2 + q^2}{2\\}$ ≥ pq

by using given equation

5 - 3pq = 2pq

5pq = 5

pq = 1

This means product of two prime number is 1 which is not possible

Hence there will be no such equation

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