Question

For any natural number n and all natural numbers d dividing 2n^2 show that n^2+d is not the square of a natural number​

Answer 1

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Answer 2

Suppose:

                                                $n^2+d=m^2;\ m\in\mathbb{N}$

Since $d$ divides $2n^2$,

                                                    $2n^2=dk;\ k\in\mathbb{N}$

Multiply the first equation by $k^2$, we get

                          $m^2k^2=n^2k^2+dk^2=n^2k^2+2n^2k=n^2(k^2+2k)$

This equation implies that $(k^2+2k)$ is a perfect square. But this is a contradiction as $k^2<k^2+2k<(k+1)^2$ and a perfect square can not lie between two consecutive squares. Hence, $(n^2+d)$ is not the square of a natural number. Q.E.D.