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Math Help - Challenging question?

  1. #1
    Super Member fardeen_gen's Avatar
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    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?
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    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

    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 again.

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

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




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