If $n$ is a positive integer such that $8n+1$ is a perfect square, then $2 n$ can't be a perfect square

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If $n$ is a positive integer such that $8n+1$ is a perfect square, then
(a) $n$ must be odd
(b) $n$ cannot be a perfect square
(c) $n$ must be a prime number
(d) $2 n$ cannot be a perfect square
Try
If we choose $n=6$ then $8n+1$ is a perfect square but $n$ is not prime and not odd . So option (a) and (c) are false. Also if we choose $n=1^2$ then $8n+1$ is a perfect square and $n$ is also a perfect square, so option (b) also false.
But how to prove option (d)

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1 Answer

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If $(8n+1)$ is a perfect square, with $n \in \Bbb{Z^+}$, then $(8n)$ can't be a perfect square.

On the other hand, if $r$ is a perfect square, then any number of the form $k^2 \times r$ also has to be a perfect square.

So, if $(2n)$ were a perfect square, then you would have to have that $(2)^2 \times (2n) = 8n$ is also a perfect square.

Therefore, the assumptions that $(8n+1)$ and $(2n)$ are both perfect squares has led to the contradictory conclusions that both $(8n+1)$ and $(8n)$ are perfect squares.

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