# A problem dealing with even perfect numbers.

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.

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Why is it that if suffices to check $p=2$ and $p=3$? There are other values of $p$ that yield perfect numbers, how do you know they won't give you $x=1$ or $x=2$? – Arturo Magidin Jul 9 '12 at 4:37
I was thinking I only need to consider the two $p$s that will give me the least residues. – HowardRoark Jul 9 '12 at 4:41
How do you know that the smallest primes will give you the smallest residues? How do you know there is no large prime $p$ such that $2^{p-2}(2^p-1)\equiv 1\pmod{5}$? A larger number may have a smaller residue: $13$ has a smaller residue modulo $5$ than $4$. – Arturo Magidin Jul 9 '12 at 4:54
I see what you are saying. Thanks. I am going to go back to work it out. – HowardRoark Jul 9 '12 at 4:57

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 $2(3)=6$.

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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)

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I don't think there are any perfect prime numbers. And The question specifically says n is and even perfect number. – HowardRoark Jul 9 '12 at 4:45
@HowardRoark: I think you misunderstand. Jeff is showing that for any odd $n$, the number $2^{n-1}(2^n-1)$ will end in either $8$ or $6$, regardless of whether $2^{n-1}(2^n-1)$ is perfect or not. So, in particular, the result will work if it just so happens that $2^{n-1}(2^n-1)$ is a perfect number (when $n$ and $2^n-1$ are both primes). Then you only need to check the case of $2(2^2-1)$, which is the only one that arises form an even exponent ($p=2$). – Arturo Magidin Jul 9 '12 at 4:53
What i am saying is p does not have to be prime, for the number to end in 6 or 8. If the power is odd , the product ends in 6 or 8 – jeff Jul 9 '12 at 4:55
@jeff oh, I misread it. I see what you guys are saying now. I got the proof. Thank you! – HowardRoark Jul 9 '12 at 5:06

$p$ 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 $8$.

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