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Quadratic Expressions, Perfect Square Not (Posted on 2010-01-03) Difficulty: 2 of 5
Prove that there cannot exist any positive integer x, such that each of 2x2 + 1, 3x2 + 1 and 6x2 + 1 is a perfect square.

See The Solution Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

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Solution Solution | Comment 5 of 6 |
Let
2x+1 = a
3x+1 = b
6x+1 = c

bc

= (6x+1)(3x+1)

= 18x^4+9x+1

= (3x)(2x+1)+1

= (3x)a+1

=> (bc) = (3xa)+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

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