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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.

Alex
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  • The main idea is correct , but the cases are not worked out properly. The product has the form $4m+3$ , correct since the product of the primes is odd. But $(1)$ $N$ need not be a prime , even if there is no prime between $4k_n+3$ and $N$ $(2)$ $N$ usually will have prime factors of the form $4l+1$, it is however crucial that there must be AT LEAST one prime factor of the form $4l+3$. – Peter Aug 20 '21 at 08:23
  • And this is the contradiction since the product is assumed to contain all the primes of the form $4l+3$ , hence $4l+3$ cannot divide $N$. So, you do not need the cases at all, you just need that not all prime factors of $N$ can have the form $4l+1$ – Peter Aug 20 '21 at 08:24
  • You can make the proof even clearer by multiplying the product of the primes with $4$ and sutracting $1$ , this makes it easier to see that $N$ must be of the form $4m+3$. – Peter Aug 20 '21 at 08:29
  • So I have to show that N is a prime of the form 4k+3. We know that any already existing 4k+3 cannot divide N, by the construction but 4k+1 may divide N. So I guess my construction of N is wrong, it would be better if I had taken all 4k+1 and 4k+3 in my consideration of N. – Alex Aug 20 '21 at 13:14
  • Can you suggest how will I ensure N is a prime, since in either case, I cannot ensure N is a prime, for eg say N could be a multiple of some 4k+1 in between largest 4kn+3 and N. – Alex Aug 20 '21 at 13:29
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    You do not need that $N$ is prime , in fact there is no known construction for arbitary large primes (which would be the case, if we could find such a product). For the contradiction it is enough to find one prime factor of $N$ of the form $4l+3$ (which cannot be in the given list, hence the list cannot be complete). – Peter Aug 20 '21 at 15:20
  • @Peter I posted my answer. I think it is correct. Could you have a look? – Alex Aug 21 '21 at 14:33
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    Remove "either" since if $N$ is prime then it has also a prime factor of the form $4k+3$ , namely $N$ itself. And perhaps you can mention that $N$ is odd , so prime factor $2$ is impossible. But apart from this , a perfect proof. – Peter Aug 21 '21 at 16:27

2 Answers2

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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.

Alex
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Your self-answer proof has logical gaps; I started editing it, but ended up just rewriting it:

Asssume that there are finitely many primes of the form $4k+3;$ let them be $p_1,p_2,\ldots,p_n.$

Let $N=4p_1 p_2 \ldots p_n-1=4(p_1 p_2 \ldots p_n-1)+3.$ Since $N$ is odd, each of its prime divisors must be of the form $4k+1$ or $4k+3.$

The identity $$(4r+1)(4s+1)=4(4rs+r+s)+1$$ shows that products of numbers of the form $4k+1$ are of the same form. Thus, since $N$ is not of this form, it has a prime divisor not of this form.

Hence, $N$ has a prime divisor $q$ of the form $4k+3,$ which is one of $p_1,p_2,\ldots,p_n.$ Therefore, $$q\mid N\quad \text {and} \quad q\mid p_1 p_2 \ldots p_n\\ q\mid(4p_1 p_2 \ldots p_n-N) \\ q\mid1,$$ i.e., $q$ is not prime. Since this is a contradiction, our assumption is false; thus, there are infinitely many primes of the form $4k+3.$

ryang
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