<s>[INST] Prove that if p and q are both prime numbers, with p > q > 2, then $p^4− q^4$ is divisible by 16 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!
 [/INST] Alternatively you may show that any odd integer power $4$ leaves the same remainder $1$ when divided by $16$:  
$(2n+1)^4 \\
= (2n)^4 + \binom{4}{1}(2n)^3 + \binom{4}{2}(2n)^2+\binom{4}{3}(2n)+1 
\\= 16(n^4+2n^3+n^2) + 8n(n+1)+1\\\equiv 1\pmod{16}$
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