PROBLEM E : 12 955 ABOUT the LACK of PERFECT SQUARES

 In GMB 5/2005 magazine, page 230 :

          E:12955  Prove that for any  $x\in\mathbb{N}$, the number  $x^2+13x+40$  

                            is not a perfect square.

([Authors:] Ioana CRĂCIUN și Gh. CRĂCIUN, Plopeni)

To avoid complicated Diophantine equation techniques, we looked for the interleaving of the number  $x^2+13x+40$  between two consecutive squares.

ANSWER CiP

$(x+6)^2<x^2+13x+40<(x+7)^2 \tag{1}$

                           Solution CiP

$0<x+4<x+13\;\;\;\;|+(x^2+12x+36)$

$\Rightarrow x^2+12x+36<x^2+13x+40<x^2+14x+49\Rightarrow (x+6)^2<x^2+13x+40<(x+7)^2$

hence (1).

$\blacksquare$