tamalkuila said:
I have proved the primes of the order 4n+3 are infiinite but can not prove that primes of the form 4n+1 are infinite.please help
prove that primes of the form 4n+1 are infinite.
This is an indirect proof by contradiction. We first
presume that only a finite number "R" of primes have form (4*n
j + 1), where {n
j, j=1,2,3,...,R, ∈ Integers}. We define integer "Q" with the following, where the product is over
all "R" primes of form (4*n
j + 1):
1: \ \ \ \ Q \ = \ \prod_{j = 1}^{R} \, (4n_{j}+1)
Now consider the integer "T" defined by the following:
2: \ \ \ \ T \ = \ 1 + Q^{2} \ = \ 1 \ + \ \left ( \prod_{j = 1}^{R} \, (4n_{j}+1) \right )^{\displaystyle 2}
There exists an odd prime "P" which divides "T", so that:
3: \ \ \ \, (1 + Q^{2}) \ \cong \ 0 \mod(P)
4: \ \ \ \ \Longrightarrow \ \ Q^{2} \ \cong \ -1 \mod(P)
Since "P" is an odd prime and gcd{P,(-1)}=(1), Eq #4 implies that (-1) is a Quadratic Residue of "P". Thus, Euler's Criterion applies:
5: \ \ \ \ (-1)^{(P - 1)/2} \ \cong \ 1 \mod(P)
Eq #5 is only true if {(P - 1)/2} is even. All odd primes have either the form (4*n + 1) or (4*n + 3). Thus, this odd prime "P" must have the form (4*n + 1) to make {(P - 1)/2} even. Hence, by Eq #1's definition of "Q", this prime "P" of form (4*n + 1) must divide "Q".
However, by definition, this "P" also divides T=(1 + Q
2):
6: \ \ \ \ P \ \textsf{\color{blue}divides} \ \left \{<br />
\begin{array}{c}<br />
Q \\<br />
\textsf{and} \\<br />
1 + Q^{2} \\<br />
\end{array}<br />
\right \} \ \ \Longrightarrow \ \ P \textsf{ \color{blue} divides } (1) \ \ \textsf{ \color{red} \LARGE ***\underline{CONTRADICTION}*** }
Thus, the original presumption of only a finite number of (4*n + 1) primes is NOT correct.
Q.E.D.
(Note: Above proof utilized Euler's Criterion for Quadratic Residues. This theorem's proof can be found at the Web Site shown below.)
http://planetmath.org/encyclopedia/ProofOfEulersCriterion.html
~~
