Problems & Puzzles:
As you know every integer of the form p^2+2pq+q^2 is a square
But what about integers of the form p^2+3pq+q^2, could it happen
to form a square integer r^2? (*)
The answer is that yes, there are many square integers that may
be expressed as p^2+3pq+q^2 = r^2.
But I only know one example where simultaneously p^2+3pq+q^2 =
r^2 and (p, q & r) are prime numbers: p=3, q=7, r=11
Q. Can you find a second example or show
that it can not exist?
(*) See pp. 82-83, Excursions in Number Theory, C. Stanley Ogilvy
& J. T. Anderson.
Solutions came from Michael Reid, Jan van Delden, Jean Brette, Emmanuel Vantieghem,
Hakan Summakoğlu, Dan Dima, Antoine Verrroken, Farideh Firoozbakht & Jahangeer
John Kholdi, & J. K. Andersen.
All of them discovered that the mentioned above solution is unique and
gave basically the same following proof (I'm using the Brette's proof):
1) case 1 : p ≠ 3
then p, q, r == 1 or 2 (mod 3) and p^2, q^2 and r^2 ==1
modulo 3 , the equation E becomes
p^2 + q^2 == r^2; which is impossible : 1+1 = 2 ≠ 1
2) case 2 : p = 3
The equation E becomes q^2 + 9q + (9 - r^2) = 0
Its discriminant is D = 81 - 4* ( 9 -
r^2) = 45 + 4 * r^2, and D must be a square k^2
So we must have 45 = k^2 - 4 * r^2 = k^2
The difference between two consecutive
squares is (n+1)^2 - n^2 = 2n+1
If 45 = 2*n + 1 ; n = 2*r = 22 and
r = 11, q = 7.
If k and 2*r are not consecutive, it is
easy to verify that there is no other solution with r < 11.