Let α and β be roots of the equation x2−nx+m=0 Given, the roots of the equation x2−nx+m=0 differ by 1 ⇒α−β=1 ⇒β=α+1 Now, The sum of the roots =α+β=−(−n)=n ⇒α+α+1=n ⇒2α+1=n ⇒α=(n−1)∕2 The product of the roots =αβ=m ⇒α(α+1)=m ⇒(
n−1
2
)(
n−1
2
+1)=m ⇒(
n−1
2
)(
n−1+2
2
)=m ⇒(n−1)(n+1)=4m ⇒n2−1=4m ⇒n2−4m−1=0 ∴ Option 1 is correct.