First of all thank you so much for answering my previous post. These are few interesting problems drawn from Prof. Gandhi lecture notes. kindly discuss:

1) If $n$ is even perfect number then $(8n +1)$ is always a perfect square.

2) Every odd perfect number has at least three different prime factors. This is by observation we can understand that. But, how to prove mathematically?

3) Every even perfect number (other than 6) can be expressible as sum of consecutive odd cubes.

Thanks in advance.

  • 1
    $\begingroup$ Even perfect numbers are triangular, and $n$ is triangular if and only if $8n+1$ is a perfect square. $\endgroup$ – Gerry Myerson Feb 5 '13 at 6:11
  • 1
    $\begingroup$ The Wikipedia link at your previous question notes that if there is an odd perfect number then it has at least nine different prime factors, and it gives a link to Nielsen, PP (2007). "Odd perfect numbers have at least nine distinct prime factors". Mathematics of Computation 76 (260): 2109–2126. doi:10.1090/S0025-5718-07-01990-4. Retrieved 30 March 2011. $\endgroup$ – Gerry Myerson Feb 5 '13 at 6:14

$1.$ By a theorem of Euler, any even perfect number is of the shape $2^{p-1}(2^p-1)$ where $p$ is prime. We don't need the primality part. Multiply by $8$, add $1$. We get $2^{2p+2}-2(2^{p+1})+1$, which is the square of $2^{p+1}-1$.

$2.$ We show that an odd perfect number cannot have only $2$ distinct prime factors (we don't deal with only $1$ prime factor, it's easier).

Let $N=p^aq^b$ where $p$ and $q$ are odd primes and $p\lt q$. Then the sum of the divisors of $N$ is $$(1+p+\cdots+p^a)(1+q+\cdots+q^b).$$ Using the ordinary formula for the sum of a finite geometric series, this can be rewritten as $$\frac{p^{a+1}-1}{p-1}\cdot \frac{q^{b+1}-1}{q-1}.$$ Divide by $N$. The result is $$\frac{p-\frac{1}{p^a}} {p-1}\cdot \frac{q-\frac{1}{q^b}} {q-1} .$$ This is less than $$\frac{p}{p-1}\cdot \frac{q}{q-1}.\tag{$1$}$$ We show that the product $(1)$ must be less than $2$. In particular, that shows the product cannot be $2$, so $N$ is in fact deficient.

It is easier to show that the reciprocal of Expression $(1)$ is greater than $\frac{1}{2}$. This reciprocal is $$\left(1-\frac{1}{p}\right)\cdot \left(1-\frac{1}{q}\right).$$

But $p\ge 3$ and $q\ge 5$. So $1-\frac{1}{3}\ge \frac{2}{3}$ and $1-\frac{1}{q}\ge \frac{4}{5}$. So their product is $\ge \frac{8}{15}$, which is greater than $\frac{1}{2}$.

I have not thought about your Question $3$.

  • $\begingroup$ Part 3 is done by knowing that it is the sum of the first $2^{(p-1)/2}$ odd cubes. $\endgroup$ – Calvin Lin Feb 5 '13 at 6:41
  • $\begingroup$ The relevant formula may not be familiar to the OP. Perhaps, if you have the time, a short answer can be given. $\endgroup$ – André Nicolas Feb 5 '13 at 7:05
  • $\begingroup$ @AndréNicolas! Thank you o much. $\endgroup$ – Jiha Feb 5 '13 at 7:46
  • $\begingroup$ @Jiha: You are welcome. $\endgroup$ – André Nicolas Feb 5 '13 at 7:50

As Andre mentioned, the even perfect numbers have the form $2^{p-1} (2^{p} -1 )$.

Claim: The sum of the first $2^{(p-1)/2}$ odd cubes is equal to this perfect number.

Proof: We know that the sum of the first $n$ odd cubes is:

$\sum_{i=1}^{n} (2i-1)^3 = \sum_{i=1}^{2n} i^3 - \sum_{i=1}^{n} (2i)^3 \\= \left[ \frac {(2n)(2n+1)}{2} \right] ^2 - 8\left[ \frac {n(n+1)}{2} \right]^2 = \frac {n^2 [16n^2 + 16n + 4 - 8(n^2 + 2n+1)]}{4}\\ = n^2(2n^2 - 1) $

Now substitute $n = 2^{(p-1)/2}$.

  • $\begingroup$ ! thank you for your explanation. $\endgroup$ – Jiha Feb 5 '13 at 7:55

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