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If $$(n^2-tn+\frac{1}{4})$$ need to be a perfect square then ,
$$(n^2-tn+\frac{1}{4})$$ = $$(n^2 + \frac{1}{\pm2}^2 - 2\times\frac{1}{\pm2}n)$$
and hence we can conclude that "t" can take ±1 value
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