Prove that there cannot exist any positive integer x, such that each of 2x² + 1, 3x² + 1 and 6x² + 1 is a perfect square.

Let
2x²+1 = a²
3x²+1 = b²
6x²+1 = c²

b²c² 

= (6x²+1)(3x²+1)

= 18x^4+9x²+1

= (3x)²(2x²+1)+1

= (3x)²a²+1

=> (bc)² = (3xa)²+1

LHS is a perfect square whereas RHS is not a perfect square for x > 0

So, the given statement is true.

Edited on January 12, 2010, 10:29 am

Posted by Praneeth on 2010-01-12 09:01:59