Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Perfect square and Positive Integer.
If n is a positive integer such that 8n+1 is a perfect square, then
Perfect square
Positive Integer
Primes
Answer: 2n cannot be a perfect square.
B.Stat Objective Problem 115
Challenges and Thrills of Pre-College Mathematics by University Press
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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