Find the values:

Euler_Euclid

Find all values of p and q if p and q are prime numbers, [tex]p^2 + q^2 + 7pq[/tex] is a perfect square.

CRGreathouse

(p, p), (3, 11), (11, 3) all work. I'd suggest trying the equation mod 4 or 36 to see if you can rule out others.

Euler_Euclid

please explain the whole solution. It was an Olympiad question and all of my friends couldn't do it.

praharmitra

[tex]p^2 + 7pq + q^2[/tex]

For p = q, it reduces to [tex]9p^2 = (3p)^2[/tex]. It therefore is a perfect square.

You can put p = 2 and show that only q = 2 gives a feasible solution, which is covered in p = q. Thus, we conclude that p,q are both odd, and hence the value of the expression is odd

For other values

[tex]
p^2 + 7pq + q^2 = (p+q)^2 + 5pq = k^2[/tex]
[tex]
5pq = k^2 - (p+q)^2 = (k - p - q)(k + p + q)
[/tex]
Thus, since p and q are primes, the only possible solutions are

[tex]
1.k-p-q = 5, k+p+q = pq
[/tex]

From the above equations we get k = 5+p+q and hence
[tex]
5+2p+2q = pq
=> 5 + 2q = p(q-2)
=> p = \frac{2q+5}{q-2} = 2 + \frac{9}{q - 2} [/tex]
[tex]
=> q - 2 = 1, 3, or 9
[/tex]

This gives (p,q) = (11,3) or (5,5) or (3,11)

The other possibilites are

[tex]
2. k-p-q = 5p, k+p+q = q [/tex](no solution)
[tex]
3. k+p+q = 5, k-p-q = pq [/tex](no solution)
[tex]
4. k+p+q = 5p, k-p-q = q [/tex](gives p = q)

Thus, the only possible solutions are (p,p), (3,11) and (11,3)

snipez90

Nice solution praharmitra. I saw the decomposition as well but didn't complete the solution since I didn't analyze the small cases at first. I have to remember that even/odd parity analysis is always a good first step in number theory equations.