A hard math question and answer often feels intimidating because the path from the problem statement to the solution is not obvious. The real challenge is not just doing calculations, but learning how to think: what to assume, what to test, what pattern to notice, and how to turn that pattern into a proof. The following problem is a classic example of a question that looks simple at first but requires a clever argument Most people skip this — try not to. That alone is useful..
People argue about this. Here's where I land on it.
A Hard Math Question: Are There Infinitely Many Primes of the Form 4k + 3?
The Question
Prove that there are infinitely many prime numbers that can be written in the form:
[ 4k + 3 ]
where (k) is a whole number.
Simply put, show that there are infinitely many primes that leave a remainder of 3 when divided by 4 Turns out it matters..
Examples include:
[ 3,\ 7,\ 11,\ 19,\ 23,\ 31,\ 43,\ 47,\dots ]
Each of these numbers gives remainder 3 when divided by 4.
This is a hard math question because it asks not merely for examples, but for a proof that the list never ends.
Understanding the Problem
Before solving it, we need to understand what the question is really asking The details matter here..
A prime number is a positive integer greater than 1 whose only positive divisors are 1 and itself. For example:
[ 2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19,\dots ]
When any integer is divided by 4, the possible remainders are:
[ 0,\ 1,\ 2,\ 3 ]
So every integer fits into one of these forms:
[ 4k,\quad 4k+1,\quad 4k+2,\quad 4k+3 ]
Prime numbers greater than 2 cannot be even, so they cannot be of the form (4k) or (4k+2). That's why, every odd prime must be either:
[ 4k+1 ]
or
[ 4k+3 ]
The question asks us to prove that the second type, (4k+3), occurs infinitely often.
The Main Idea of the Proof
The proof uses a famous technique called proof by contradiction The details matter here..
In a proof by contradiction, we begin by assuming the opposite of what we want to prove. Then we show that this assumption leads to something impossible.
Here, we want to prove:
There are infinitely many primes of the form (4k+3).
So we assume the opposite:
There are only finitely many primes of the form (4k+3) But it adds up..
If that assumption leads to a contradiction, then the original statement must be true.
Step-by-Step Solution
Step 1: Assume There Are Only Finitely Many Such Primes
Suppose there are only finitely many primes of the form (4k+3).
Let us list them:
[ p_1,\ p_2,\ p_3,\dots,\ p_n ]
Each of these primes satisfies:
[ p_i \equiv 3 \pmod{4} ]
Here's one way to look at it: if we only knew a few such primes, the list might begin:
[ 3,\ 7,\ 11,\ 19,\ 23,\dots ]
But under our assumption, this list is complete.
Step 2: Build a Special Number
Now consider the number:
[ N = 4p_1p_2p_3\cdots p_n - 1 ]
This number is carefully designed.
First, notice that:
[ N = 4(p_1p_2p_3\cdots p_n) - 1 ]
Since (4(p_1p_2p_3\cdots p_n)) is divisible by 4, subtracting 1 means:
[ N \equiv -1 \pmod{4} ]
But (-1 \equiv 3 \pmod{4}), so:
[ N \equiv 3 \pmod{4} ]
So, (N) is also of the form (4k+3) Worth keeping that in mind..
Step 3: Check Whether Any Listed Prime Divides (N)
Now divide (N) by any prime (p_i) from our list.
Since:
[ N = 4p_1p_2p_3\cdots p_n - 1 ]
we have:
[ N \equiv -1 \pmod{p_i} ]
That means when (N) is divided by (p_i), the remainder is (-1), or equivalently (p_i - 1).
So none of the primes (p_1, p_2, p_3,\dots,p_n) divides (N).
This is important Simple, but easy to overlook..
We constructed (N) so that it is not divisible by any known prime of the form (4k+3).
Step 4: Factor (N) Into Primes
Every integer greater than 1 can be written as a product of prime numbers. This is known as the Fundamental Theorem of Arithmetic.
So (N) has a prime factorization.
Take this: if (N) were 91, we could write:
[ 91 = 7 \times 13 ]
In our case, we do not need to know the exact factorization of (N).
