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.
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
$