Find all pairs of primes ## p ## and ## q ## satisfying ## p-q=3 ##.

[Math100]

Homework Statement

Find all pairs of primes $p$ and $q$ satisfying $p-q=3$.

Relevant Equations

None.

Proof:

Let $p$ and $q$ be primes such that $p-q=3$.
Now we consider two cases.
Case #1: Suppose $p$ is an even prime.
Then $p=2$, because $2$ is the only even prime.
Thus $2-q=3$, so $q=-1$,
which contradicts the fact that $q$ is prime.
Case #2: Suppose $p$ is an odd prime.
Then $p=2k+1$ for some $k\in\mathbb{N}$.
Thus $2k+1=q+3$
$q=2k+1-3$
$=2k-2$
$=2(k-1)$.
This means $q$ is an even prime.
Now we have $p-2=3$, so $p=5$.
Therefore, the pair of primes $p$ and $q$ satisfying $p-q=3$ is $(p, q)=(5, 2)$.

 

[Mark44]

Homework Statement:: Find all pairs of primes $p$ and $q$ satisfying $p-q=3$.
Relevant Equations:: None.

Proof:

Let $p$ and $q$ be primes such that $p-q=3$.
Now we consider two cases.
Case #1: Suppose $p$ is an even prime.

This is much more general than it needs to be. The only even prime is 2, so the only primes that satisfy this case are q = 2 and p = 5.
Better:
$p-q=3 \Rightarrow p = q + 3$
By inspection it can be seen that q = 2 and p = 5 are two primes that satisfy the given equation.

Now suppose that q is prime with q > 2.
Necessarily q is odd, with q = 2n + 1 for some integer n.
Then p = q + 3 = 2n + 1 + 3 = 2n + 4 = 2(n + 2), which is an even integer, this not a prime.
$\Rightarrow \Leftarrow$ Note that a pair of arrows that butt heads are often used to indicate a contradiction.
Therefore, the only primes that differ by 3 are 2 and 5.

Then $p=2$, because $2$ is the only even prime.
Thus $2-q=3$, so $q=-1$,
which contradicts the fact that $q$ is prime.
Case #2: Suppose $p$ is an odd prime.
Then $p=2k+1$ for some $k\in\mathbb{N}$.
Thus $2k+1=q+3$
$q=2k+1-3$
$=2k-2$
$=2(k-1)$.
This means $q$ is an even prime.
Now we have $p-2=3$, so $p=5$.
Therefore, the pair of primes $p$ and $q$ satisfying $p-q=3$ is $(p, q)=(5, 2)$.

 

[pbuk]

If $p - q = 3$ then $p$ and $q$ have opposite parity.
The only even prime is $2$ so either $p = 2$ or $q = 2$.
$p = 2 \implies q = -1$ which is not a solution.
$q = 2 \implies p = 5$ which is therefore the only solution.