Perfect square and Positive Integer | TOMATO B.Stat Objective 115

If n is a positive integer such that 8n+1 is a perfect square, then

Let \((8n+1)=k^{2}\) be a perfect square so k is found to be odd here as 8n+1 is odd.

\(\Rightarrow 8n=k^{2}-1\)

\(\Rightarrow 8n=(k-1)(k+1)\)

\(\Rightarrow 2n=\frac{(k-1)(k+1)}{4}\)

\(\Rightarrow (\frac{k-1}{2})(\frac{k+1}{2})\)

here (k-1) and (k+1) are consecutive even numbers then \((\frac{k-1}{2})(\frac{k+1}{2})\) are consecutive even numbers

\(2k \times (2k+2)= 2^2{k(k+1)}\) is not a perfect square as product of two consecutive numbers proved here as not a perfect square

So, 2n is not perfect square.

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