A problem dealing with even perfect numbers.

Question

Question: Show that all even perfect numbers end in 6 or 8.

This is what I have. All even perfect numbers are of the form $n=2^{p-1}(2^p -1)$ where $p$ is prime and so is $(2^p -1)$.

What I did was set $2^{p-1}(2^p -1)\equiv x\pmod {10}$ and proceeded to show that $x=6$ or $8$ were the only solutions.

Now, $2^{p-1}(2^p -1)\equiv x\pmod {10} \implies 2^{p-2}(2^p -1)\equiv \frac{x}{2}\pmod {5}$, furthermore there are only two solutions such that $0 \le \frac{x}{2} < 5$. So I plugged in the first two primes and only primes that satisfy. That is if $p=2$ then $\frac{x}{2}=3$ when $p=3$ then $\frac{x}{2}=4$. These yield $x=6$ and $x=8$ respectively. Furthermore All solutions are $x=6+10r$ or $x=8+10s$.

I would appreciate any comments and or alternate approaches to arrive at a good proof.

Answer 1

Note that the powers of $2$ are congruent to $2$, $4$, $8$, or $6$, according to whether the exponent is congruent to $1$, $2$, $3$, or $0$ modulo $4$.

Assume $p\equiv 1\pmod{4}$. Then $2^{p-1}\equiv 6\pmod{10}$, and $2^p-1\equiv 1\pmod{10}$, so the product is congruent to $6$ modulo $10$.

If $p\equiv 3\pmod{4}$, then $2^{p-1}\equiv 4\pmod{10}$, and $2^p-1\equiv 7\pmod{10}$, so the product is congruent to $8$ modulo $10$.

Finally, if $p=2$, then $2(3)=6$.

Answer 2

In this proof, n is any integer(not just primes)

The number 2^(n-1) last digit repeats in the four number cycle(2,4,8,6)
The number 2^(n) -1 last digit repeats in the four number cycle (3,7,5,1)
The product's last digit repeats in the four number cycle (6,8,0,6)

If n is odd, the pattern repeats (8,6)

You will find the number end in 6 or 8 for all odd numbers(not just primes)

Answer 3

$p$ is prime so it is 1 or 3$\mod 4$.

So, the ending digit of $2^p$ is (respectively) 2 or 8 (The ending digits of powers of 2 are $2,4,8,6,2,4,8,6,2,4,8,6...$

So, the ending digit of $2^{p-1}$ is (respectively) 6 or 4; and

the ending digit of $2^p-1$ is (respectively) 1 or 7.

Hence the ending digit of $2^{p-1}(2^p-1)$ is (respectively) $6\times1$ or $4\times7$ modulo 10, i.e., $6$ or $8$.