Wrong with the proof of there is no odd perfect number

Question

The Conjecture:

$A \equiv_2 1$ Implies $\sigma(A) \not = 2A$

The Proof:

Let $H :=$ $[A = 2Q + 1]$ $\land$ $[\sigma(A) = 2A]$. So $H$ is true IFF

$$\sigma(2Q + 1) = 2(2Q + 1).$$

Therefore we will show this is impossible by contradiction and it is going to imply our conjecture is true. Let me remind few pieces of information for sake of clarity in my proof.

1) We can calculate $\sigma(P^K)$ for prime $P$ by the sum:

$$\sigma(P^K) = \sum_{I=0}^K{P^I}$$

2) The $\sigma$ is a multiplicative function: $\sigma(PQ) = \sigma(P)\sigma(Q)$ for $(P, Q) = 1$. Hence we can calculate $\sigma$ for any composite number by the unique prime factorization property.

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$$\sigma(2Q + 1) = 2(2Q + 1) \equiv_2 0$$

So if we factorize $A = P_1^{K_1}P_2^{K_2}…$ for calculating $\sigma$:

$$\sigma(A) = \prod_{I=1}\sum_{J=0}^{K_I} P_I^{J} \equiv_2 0$$

Hence one of the factors must be divisible by $2$:

$$\sum_{J=0}^{K_I} P_I^{J} = 1 + P_I^1 + P_I^2 + … + P_I^{K_I} \equiv_2 0$$
$$P_I^1 + P_I^2 + … + P_I^{K_I} \equiv_2 1$$

Since all $P_I$ are odd, then the $K_I$ must be odd. Since there are $K_I$ odd terms, even $K_I$ would leave us with an even sum.

$$K_I \equiv_2 1.$$

But this means that $A$ must not be some perfect square otherwise it's factorization can't contain a prime raised to some odd exponent,

$$A \not = X^2, \forall X \in N.$$

Hence,

$$A \not \equiv_2 X^2, \forall X \in N.$$

But this would mean $A$ is a Quadratic Non-Residue under modulo $2$. But yet we know the opposite is true since $A \equiv_2 1$ and $1$ is a Quadratic Residue modulo $2$.

Q.E.D

Answer 1

Your main problem is:

• You don't seem to realize that not just squares occupy the quadratic residues. If they are coprime to the modulus, they can also encode possible primes, for example. Certain coprime residues, can only be primes in certain multipliers of the modulus. For example, 2 or 4 mod 5, will never be prime within even multipliers of 5, as the result will be even. The residue 4, will never return a prime when in a 1 mod 3 multiplier of 5. The residue 2, will never give a prime in 2 mod 3 multipliers of 5 etc.

Other problems include:

• You don't know what square-free (originally claimed) means, it means not having any exponents greater than 1 in the prime factorization. They can still land in quadratic residue classes.

• Odd perfect numbers, have a known form if they exist. All you've really managed, is that the two separate parts, don't have all the same primes ( or congruent to the same primes) raised to same parity exponents.