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1) If $p ($$2^r$$/2$) is an even perfect number then $p$ should be in the form of $2^r$ - 1

2) Every even perfect number ends with 6 or 28. why/justify

3) If $a^k$ - 1 is prime for $a > 0$ and $k\ge 2$. If we fix $a = 2$, then $k$ becomes prime. Also, in what cases of a, we get primes always?

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$1$. For the first one, note that $$\sum_{d \vert p \cdot 2^{r-1}}d = 1 + 2 + 2^2 + \cdots 2^{r-1} + p(1 + 2 + 2^2 + \cdots 2^{r-1}) = 2^r-1 + p (2^r-1) = (p+1)(2^r-1)$$ Since we want it to be perfect number, we have $$(p+1)(2^r-1) = 2 \times p \cdot 2^{r-1} = p \cdot 2^r \implies p \cdot 2^r - p + 2^r - 1 = p \cdot 2^r$$ Hence, $$p = 2^r - 1$$


$2$. For the second part, it is not hard to show that all even perfect numbers are of the form $$(2^r-1) \cdot 2^{r-1}$$ such that $2^r-1$ is a prime. Since $2^r-1$ is a prime, $2^r-1$ ends in $1,3$ or $7$. Hence, $2^r \equiv 2,4,8 \pmod{10} \implies 2^{r-1} \equiv 6,2,4 \pmod{10}$. Hence, $$2^{r-1} \cdot (2^r-1) \equiv 6 \times 1, 2 \times 3, 4 \times 7 \pmod {10} \equiv 6,8 \pmod{10}$$ Hence, even perfect number ends in $6$ or $8$. If it ends in $8$, we want to show that it in fact ends in $28$. We have that if $2^r \equiv 8 \pmod{10}$, then $r \equiv 3 \pmod4$. Hence, $$2^{2r-1} - 2^{r-1} = 2^{8k+5} - 2^{4k+2} = 32 \cdot 256^k - 4 \cdot 16^k$$ Now we have $$256^k \equiv \begin{cases}56 \pmod {100} & k \equiv 1 \pmod 5\\ 36 \pmod {100} & k \equiv 2 \pmod 5\\ 16 \pmod {100} & k \equiv 3 \pmod 5\\ 96 \pmod {100} & k \equiv 4 \pmod 5\\ 76 \pmod {100} & k \equiv 0 \pmod 5 \end{cases}$$ and $$16^k \equiv \begin{cases}16 \pmod {100} & k \equiv 1 \pmod 5\\ 56 \pmod {100} & k \equiv 2 \pmod 5\\ 96 \pmod {100} & k \equiv 3 \pmod 5\\ 36 \pmod {100} & k \equiv 4 \pmod 5\\ 76 \pmod {100} & k \equiv 0 \pmod 5 \end{cases}$$ Hence,$$32 \cdot 256^k - 4 \cdot 16^k \equiv 28 \pmod{100} \,\,\,\,\,\, \forall k \in \mathbb{Z}^+$$ Hence, even perfect numbers end in $6$ or $8$. If it ends in $8$, then it in-fact ends in $28$.


$3$. For the last part, if $2^k-1$ is a prime, then $k$ has to be a prime. This can be proved easily by looking at the contrapositive of this statement. If $k$ is composite, then $k = m \times n$, where $m,n > 1$. This gives us $$2^k - 1 = 2^{mn} - 1 = (2^m)^n - 1 = (2^m-1)(1+2^m + 2^{2m} + \cdots + 2^{(n-1)m})$$ Hence, $2^m-1 > 1$ divides $2^{mn} - 1$. Hence, $2^k-1$ is composite if $k$ is composite i.e. to put it the other way, if $2^k-1$ is a prime ,then $k$ is a prime. Also, note that $a^k-1$ is never a prime for $a>2$, since $$a^k-1 = (a-1) (1+a+a^2 + \cdots a^{k-1})$$ and since $a-1 > 1$, we have that $a^k-1$ is composite (for $a>2)$, irrespective of whether $k$ is a prime or not.

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