Prove that if p and q are both prime numbers, with p > q > 2, then $p^4− q^4$ is divisible by 16

Question

Prove that if p and q are both prime numbers, with p > q > 2, then $p^4 − q^4$ is divisible by 16.

This is my attempt so far:

Since p and q are both prime numbers greater than 2, then they must be odd and hence can be written:

$p=2m+1$, for some integer n,

And $q=2n+1$, for some integer m.

Consider, $p^4-q^4$ $\implies$ $(p^2+q^2)(p+q)(p-q)$

$p-q=2m+1-(2n+1)=2(m-n),$ hence is a multiple of 2.

$p+q=2m+1+2n+1=2(m+n+1),$ hence is a multiple of 2.

$p^2+q^2=(2m+1)^2+(2n+1)^2=$

$2(2m^2+2m+2n^2+2n+1),$ hence is a multiple of 2.

However, this is where I am stuck. Doesn’t this just prove the statement is a multiple of 8, not 16? Could anyone help with this proof, or maybe suggest an alternative method. Thanks!

Answer 1

Also $(m-n)$ or $(m+n+1)$ is a multiple of $2$

Answer 2

Alternative proof:

If $p$ is odd, then $8|p^2-1$ and $2|p^2+1$, so $8\times2=16|(p^2-1)(p^2+1)=p^4-1$.

Likewise, $16|q^4-1$. Therefore $16|(p^4-1)-(q^4-1)=p^4-q^4.$

Answer 3

We can prove something more generic

For any odd number $n=2m+1,$

$n^2=8\cdot\dfrac{m(m+1)}2+1=8r+1$(say) where $r$ is an integer

$n^4=(1+8r)^2=1+16r+64r^2\equiv1\pmod{16}$

Can you take it from here?

Answer 4

You're off to a great start.

Rewrite $$p^4-q^4=(p^2+q^2)(p+q)(p-q).$$

Regarding $p^2+q^2$:

Since $p$ and $q$ are both odd, they can be written as $p=2m+1$ and $q=2n+1$. Then we have \begin{align} p^2+q^2&=(2m+1)^2+(2n+1)^2\\ &=4m^2+4m+1+4n^2+4n+1\\ &=2(2m^2+2m+2n^2+2n+1) \end{align} This tells us that $2|p^2+q^2$ (and is exactly what you have).

Regarding $(p+q)(p-q)$:

Odd numbers are either $1\pmod 4$ or $3\pmod 4$. We have three cases:

Case 1: let $p\equiv1\pmod4$ and $q\equiv1\pmod 4$. Then we have $p+q\equiv 2\pmod 4$ and $p-q\equiv 0\pmod 4 \implies 4| p-q$.

Case 2: WLOG, let $p\equiv3\pmod4$ and $q\equiv1\pmod4$. Then we have $p+q\equiv0\pmod 4\implies 4|p+q$ and $p-q\equiv2\pmod4$.

Case 3: let $p\equiv3\pmod4$ and $q\equiv3\pmod4$. Then we have $p+q\equiv2\pmod4$ and $p-q\equiv0\pmod4\implies 4|p-q$.

In each of these three cases, while both $p+q$ and $p-q$ are both divisible by $2$, always only one of them is divisible by $4$. As such, $8|p^2-q^2$ for odd $p$ and $q$.

Together, as we have $2|p^2+q^2$ and $8|p^2-q^2$, we have $16|(p^2+q^2)(p^2-q^2)\implies 16|p^4-q^4$ as desired.

Answer 5

Alternatively you may show that any odd integer power $4$ leaves the same remainder $1$ when divided by $16$:

$(2n+1)^4 \\ = (2n)^4 + \binom{4}{1}(2n)^3 + \binom{4}{2}(2n)^2+\binom{4}{3}(2n)+1 \\= 16(n^4+2n^3+n^2) + 8n(n+1)+1\\\equiv 1\pmod{16}$