Show that even perfect numbers (other than 6) have a remainder 1 on division by 9 Every even perfect number can be written in the form, $2^{p-1}(2^p-1)$ so this means I want to show that $2^{p-1}(2^p-1)\equiv1$ mod $9$ but can't see where to go from here.
 A: For even perfect numbers other than 6, we know that $p$ is odd. Powers of $2$, modulo $9$, are congruent to $1, 2, 4, 8, 7, 5$, as the exponent is congruent to $0,1,2,3,4,5$ mod $6$, respectively. That gives us that $2^{p-1}$ and $2^p$ are congruent, modulo $9$, to $1$ and $2$, or $4$ and $8$, or $7$ and $5$. In any case, the product $2^{p-1}(2^p-1)$ comes out to $1$ modulo $9$.
A: Probably not the most beautiful approach but sure is a quick one:
$2^{n}$ is congruent to $\{2,4,-1,-2,-4,1\}$ mod$(9)$.
Hence, for a prime number $p$, $2^p$ is congruent to $2$ or $-4$ mod $(9)$. In both cases it's easy to check that $2^{p-1}(2^p-1)$ is congruent to $1$ modulo $9$
A: We have $n=(2^p-1)(2^{p-1})$.  Since the Euler totient of $9$ is $6$, consider possible residues of $p$ modulo $6$.  For perfect numbers greater than $6$ this residue must be $1$, $3$ or $5$.
If $p\equiv 1 \pmod 6$ then $2^p-1 \equiv 1 \pmod 9$ and $2^{p-1} \equiv 1 \pmod 9$. Thus their product has residue $1$ modulo $9$.  The reader is left with the other cases.
A: Let $x=2^{p-1}$.  To show that $2^{p-1}(2^p-1)=2x^2-x\equiv1$ mod $9$ it suffices to note that
$$2x^2-x-1=(2x+1)(x-1)$$
and observe that $p-1$ is even since $p\gt2$, which implies $x=2^{p-1}\equiv1$ mod $3$, so that $3\mid2x+1$ and $3\mid x-1$.
