Proof: Twin Primes Always Result in Perfect Squares

[Math100]

Homework Statement

If $1$ is added to a product of twin primes, prove that a perfect square is always obtained.

Relevant Equations

None.

Proof:

Suppose $p$ and $p+2$ are twin primes.
Then we have $p(p+2)+1=p^2+2p+1=(p+1)^2$.
Thus, $(p+1)^2$ is a perfect square.
Therefore, if $1$ is added to a product of twin primes,
then a perfect square is always obtained.

 

[fishturtle1]

Looks good to me. Stylistically, one might write "Thus, $p(p+2)+1$ is a perfect square." instead of "Thus, $(p+1)^2$ is a perfect square.".

 

[Orodruin]

… although the restriction to twin primes is unnecessary. As should be clear from the proof, it holds for any two integers that differ by two.

Edit: There is also a rather intuitive geometric interpretation: Make a square out of (p+1)^2 unit boxes. Take the top row containing p+1 unit boxes and place p of them in a column on the right side of the rectangle, thus leaving you with a rectangle of side lengths p and p+2 with a single leftover unit box.

Edit 2: Illustration

Edit 3: Even easier to see with $q = p+1$, i.e.,
$$
(q-1)(q+1) = q^2 - 1 \quad \Leftrightarrow \quad q^2 = (q-1)(q+1) + 1.
$$

Edit 4: The generalisation being cutting a strip of width b from the top of a square of side length a and placing a portion of length a-b of the strip to the right of the square, leaving a rectangle with sides a-b and a+b and a square of side length b:
$$
a^2 = (a-b)(a+b) + b^2.
$$
Regardless of $a$ and $b$ being integers or not.

 

[pbuk]

Stylistically, one might write "Thus, $p(p+2)+1$ is a perfect square." instead of "Thus, $(p+1)^2$ is a perfect square.".

This is more than a stylistic point: if you say "Thus, $(p+1)^2$ is a perfect square" then you are saying "because $p(p+2)+1=p^2+2p+1=(p+1)^2$ then $(p+1)^2$ is a perfect square" which is not correct (note 1). The words "Then we have" are also not appropriate here because that is saying "because $p$ and $p+2$ are twin primes then $p(p+2)+1=p^2+2p+1$" which is not correct (note 2).

The proof simply needs to be:

Suppose $p$ and $p+2$ are twin primes.
$p(p+2)+1=p^2+2p+1=(p+1)^2$, which is a perfect square.

Note 1: $(p+1)^2$ is a perfect square independent of the fact that $p(p+2)+1=p^2+2p+1=(p+1)^2$.
Note 2: $p(p+2)+1=p^2+2p+1$ independent of whether $p$ and $p+2$ are twin primes.