The greatest common divisor of two integers YOU and ME is the greatest integer that divides both YOU and ME......REMEMBER THE NUMBER THEORY

Note that every integer has positive and negative divisors. If a is a positive
divisor of mE, then −YOU is also a divisor of mE. Therefore by our definition of the
greatest common divisor, we can see that nobody wants to be .....you ....and nobody dares to be me ...

More on the Infinitude of Primes
There are also other theorems that discuss the infinitude of primes in a given arithmetic
progression. The most famous theorem about primes in arithmetic progression
is Dirichlet’s theorem
Theorem 16. Dirichlet’s Theorem Given an arithmetic progression of terms an+
b , for n = 1, 2, ... ,the series contains an infinite number of primes if a and b are
relatively prime,
This result had been conjectured by Gauss but was first proved by Dirichlet.
Dirichlet proved this theorem using complex analysis, but the proof is so challenging.
As a result, we will present a special case of this theorem and prove that
there are infinitely many primes in a given arithmetic progression. Before stating
the theorem about the special case of Dirichlet’s theorem, we prove a lemma that
will be used in the proof of the mentioned theorem.
Lemma 7. If a and b are integers both of the form 4n + 1, then their product ab
is of the form 4n + 1
Proof. Let a = 4n1 + 1 and b = 4n2 + 1, then
ab = 16n1n2 + 4n1 + 4n2 + 1 = 4(4n1n2 + n1 + n2) + 1 = 4n3 + 1,
where n3 = 4n1n2 + n1 + n2.
Theorem 17. There are infinitely many primes of the form 4n + 3, where n is a

positive integer