Show that gcd(9t+4, 2t+1)=1 for all integers tElementary number theory

The gcd of two numbers a and b is equal to
(a*b)/lcm(a,b)

For 9t+4 and 2t+1, we
have

gcd(9t+4, 2t+1) =
(9t+4)(2t+1)/lcm((9t+4),(2t+1))

9t+4 = 4(2t+1)+t. 4(2t+1)+t
is a multiple of 2t+1 only when 2t+1 = t or t = -1. When t = -1, 9t+4 = -5 and 2t+1=-1.
-5 and -1 have a gcd of 1. If 2t+1 is not equal to t, 4(2t+1)+t cannot be a multiple of
2t+1. The lcm of (9t+4) and (2t+1) is therefore equal to
(9t+4)(2t+1)

The value of the gcd of the two terms is given
by (9t+4)(2t+1)/(9t+4)(2t+1) = 1

This proves
that the gcd of (9t+4) and (2t+1) is 1.