Thread: Infintely many primes of the form:

1. Infintely many primes of the form:

Hi,

Prove that there exists infinetly many primes of the form:

a) 4q + 1 for some q in N U {0}
b) 4q + 3 for some q in N U {0}

Any help would be greatly appreciated.

2. a) consider the polynomial $\displaystyle f(x) = x^2 + 1$

Fix $\displaystyle x$ and suppose $\displaystyle {f\left( x \right)}$ has a prime divisor $\displaystyle p>2$ then $\displaystyle x^2 \equiv - 1\left( {\bmod .p} \right)$

By Fermat's Little Theorem: $\displaystyle 1\equiv{x^{p-1}}= (x^2)^{\tfrac{p-1}{2}} \equiv (-1)^{\tfrac{p-1}{2}}\left( {\bmod .p} \right)$ and so $\displaystyle \left. 4 \right|\left( {p - 1} \right)$ that is $\displaystyle p \equiv 1\left( {\bmod .4} \right)$ (1)

Now assume there were finitely many primes $\displaystyle \equiv 1 (\bmod.4)$ let them be $\displaystyle \theta _1 ,...,\theta _n$ and consider $\displaystyle f( 2\cdot \theta _1 \cdot ... \cdot \theta _n )$ you can check that $\displaystyle f\left( {2\cdot \theta _1 \cdot ... \cdot \theta _n } \right) \equiv 1\left( {\bmod .\theta _i } \right)$ $\displaystyle \forall i \in \left\{ {1,...,n} \right\}$ and so it is not divisible by a prime $\displaystyle \equiv{1}(\bmod.4)$ however this contradicts (1), since $\displaystyle f(2\cdot \theta _1 \cdot ... \cdot \theta _n )$ must have at least one odd prime divisor (since it's odd>1). $\displaystyle \square$

b) This one is easier. Suppose there were finitely many $\displaystyle \rho _1 ,...,\rho _n$, and consider $\displaystyle 4 \cdot \left( {\rho _1 \cdot ... \cdot \rho _n } \right) + 3$

3. Thanks alot.

I understand.