Challenging question?

• July 1st 2008, 01:05 AM
fardeen_gen
Challenging question?
Find all primes p and q such that p^2 + 7pq + q^2 is the square of an integer?

The answer is not important. What is the approach to this problem?
• July 1st 2008, 03:34 AM
Isomorphism
Quote:

Originally Posted by fardeen_gen
Find all primes p and q such that p^2 + 7pq + q^2 is the square of an integer?

The answer is not important. What is the approach to this problem?

First of all, with $q = p, p^2 + 7pq + q^2 = 9p^2 = (3p)^2$, so any prime would do.

So from now on lets assume $p\neq q$

Let $p^2 + 7pq + q^2 = x^2 = (p+q)^2 + 5pq \Rightarrow x^2 - (p+q)^2 = (x - p - q)(x+p+q)= 5pq$

Now since $5,p,q$ are all primes, the factors of $5pq$ are $1,5pq,5p,5q,pq,p,q,5$

Observing that $x- p - q < x+p+q$, we have

1) $x-p-q = 1, x+p+q = 5pq \Rightarrow 2(p+q)+1 = 5pq \Rightarrow q = \frac{2p+1}{5p-2}$

Assume a solution exists. Now since q is a prime, $q > 1 \Rightarrow 2p+1 > 5p - 2 \Rightarrow p < 1$, which is not possible since we assumed p to be prime and hence p > 1. Thus no solution exists.

2) $x-p-q = q, x+p+q = 5p \Rightarrow 2p + 3q = 5p \Rightarrow q=p$

We have already solved this case.

3) $x-p-q = p, x+p+q = 5q \Rightarrow 3p + 2q = 5q \Rightarrow q=p$

Same as (2)

4) $x-p-q = 5, x+p+q = pq \Rightarrow 5 + 2(p + q) = pq \Rightarrow q = \frac{2p+5}{p-2} = 2 + \frac{9}{p-2}$

Since q is a natural number, (p-2)|9 and thus p-2 can be either 1,3 or 9.

So trying all cases,

1) $p-2=1 \Rightarrow p=3 \Rightarrow q = 2 + \frac{9}{p-2} = 11$

Thats (Rock)

2) $p-2=3 \Rightarrow p=5 \Rightarrow q = 2 + \frac{9}{p-2} = 5$

Its right but we already have tackled $p=q$ case.

3) $p-2=9 \Rightarrow p=11 \Rightarrow q = 2 + \frac{9}{p-2} = 3$
Thats (Rock) again.

So the solutions are the following(note that w represents any arbitrary prime):

$(p,q) \in \{(w,w),(3,11),(11,3)\}$