Problem
NT-B2-M01-P005 A Square and an Odd Divisor
#5
★★☆☆☆ Level 2 of 5
Prove that \( \gcd(2n+1,4n^2+4n+3)=1 \) for every integer \(n\).
Compare the second number with the square of \(2n+1\).
We have \((2n+1)^2=4n^2+4n+1\). If \(d\) divides \(2n+1\) and \(4n^2+4n+3\), then \(d\) divides the difference \(2\). But \(2n+1\) is odd, so the common divisor cannot be \(2\). Hence \(d=1\).
This problem trains recognition of an almost-square.