We shall prove this theorem by contradiction. Assume $p_n$ is the largest prime number of the form 4$k_n$+3 for some n $\in$ $\mathbb{N}$. Consider an odd value of n. Even if n is even, we could miss one of the prime numbers to get an odd n. Now, consider
\begin{align*}
N &= 2(4 k_1 +3)(4 k_2 +3)(4k_3+3)…(4 k_n +3) + 1 \\
&= 4m + 2\cdot 3^n + 1 \\
&= 4m + 2\cdot (4-1)^n + 1 \\
&= 4m + 2\cdot 4a -2 + 1 \\
&= 4m + 2\cdot 4a -1 \\
&= 4m + 3
\end{align*}
for some m,a $\in$ $\mathbb{N}$
Two cases are possible. N is definitely not divisible by 4$k_i$+3 i $\in$ [1,n].
- All numbers between 4$k_n$+3 and N are composite $\implies$ they are formed from a unique prime factorisation of the finite set of prime numbers. Then N is prime. so contradiction.
- There is a new prime between 4$k_n$+3 and N of the form 4k+3 so N is a multiple of that prime(Note that N cannot be a multiple of 4k+1 since we have 2 occurring only once in our expression for N-1), which any way is again a contradiction.
My prof said that this proof misses some cases but I am unable to figure out which case. Please help. I feel pretty confident of my proof.
Best Answer
We shall prove this theorem by contradiction. Assume $p_n$ is the largest prime number of the form 4$k_n$+3 for some n $\in$ $\mathbb{N}$. Now, consider \begin{align*} N &= 4(4 k_1 +3)(4 k_2 +3)(4k_3+3)...(4 k_n +3) - 1 \\ &=4m+3 \end{align*} for some m $\in$ $\mathbb{N}$
N is definitely not divisible by any of 4$k_i$+3 i $\in$ [1,n].
Now either N is prime or N has at least one prime factor of the form 4$k$+3 apart from itself, not in this list since if N had all prime factors of the form 4k+1, then N would have been of the form (4$u_1$+1)(4$u_2$+1)(4$u_3$+1)...(4$u_n$+1) = 4l+1 only. so a contradiction. Hence proved.
I accommodated some of Peter's changes, hope this proof is correct.