Problem
NT-B1-M12-P015 Set 3. Divisor \(2n-1\)
#15
★★★☆☆ Level 3 of 5
Find all positive integers \(n\) such that \(2n-1\mid n^2+1\).
Multiply by \(4\).
Let \(d=2n-1\). If \(d\mid n^2+1\), then \(d\mid4(n^2+1)\). But \(4(n^2+1)=d^2+2d+5\). Thus \(d\mid5\). Hence \(d=1\) or \(5\), giving \(n=1\) or \(n=3\). Both work.
Useful template with a linear divisor.