Problems & Puzzles: Puzzles

 Puzzle 623. p^2+3pq+q^2=r^2 As you know every integer of the form p^2+2pq+q^2 is a square integer r^2. 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.

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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 (mod 3)
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 - (2*r)^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.

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