1we have
p2 + q2 +7pq = k2
k2 - (p + q)2 = 5pq
(k + p + q)(k - p - q) = 5pq
CASE 1 ; k - p -q = 5 , pq = k + p +q
pq = 2 (p + q) + 5
p = (2q + 5)/(q-2) = 2 + 9/(q-2)
9/q-2 must be an integer so q=3,5,11 p = 11,5,3
CASE 2 :k + p+q = 5 , pq = k - p - q
and similarly you can see other cases
21THNKS for the solution
(i assumed there can't be any solution and was trying to find contradictions only by congruences ;(
1
fibonacci
·2011-05-12 01:58:09
also p=q=any prime are more solutions :):)
21
Shubhodip
·2011-05-12 02:22:22
omg i missed even that !!!