# Is this proof that there are no perfect, odd, integer square numbers legitimate?

### Assumptions:

Any even number times any other number is always an even number. An odd number times an odd number is always an odd number. An even number plus an even number is even, and an odd number plus an odd number is odd. An even number plus an odd number is odd. An even number is not odd.

A perfect number is one such that the sum of its factors is equal to itself, not including itself.

Conjecture: There are no odd, perfect numbers.

Let $n$ be a odd, perfect,square number. Its factors can be represented as a list: $$a,b,c ..... ,\sqrt n ,....$$ Every factor must be odd, if our assumptions are correct. (Odd * Odd = Odd) Let us call any factor of $n$, $F$. There is another number that when multiplied with $F$ that results in $n$. $$1\cdot n = n$$ $$a \cdot a' = n$$ $$b \cdot b' = n$$ ... $$\sqrt n \sqrt n = n$$

Take the sums of these pairs, except for $\sqrt n \sqrt n$ and $1\cdot n$. You get: $$E + E' + E''......$$

Since any odd number plus another odd is even. Summing those numbers, we get some number $T$. Now, we see that the 2 remaining factors*, $\sqrt n$, and $1$, sum to an even number. (Odd + Odd is Even) Combining these, we get an even number.

*: $\sqrt n$ is the same as $\sqrt n$. $n$ is not considered a factor.

Since a even number $(T + \sqrt n +1)$ cannot equal an odd number $(n)$, we have proven there are no odd, perfect, square numbers.

• Whoops, misread. Thanks, @900sit-upsaday – Thomas Andrews Aug 13 '14 at 3:52
• The proof is correct, but it would look better with some LaTeX formatting. See math notation guide. – user147263 Aug 13 '14 at 3:52
• Nice proof (+1). Now try proving that there are no odd perfect numbers that are perfect powers ;) – Prism Aug 13 '14 at 3:59
• "an odd number plus an odd number is odd" - incorrect. – James Aug 13 '14 at 17:31

Yes, your proof is correct. You have shown that there are no odd perfect numbers that are squares.

The conjecture that there are no odd perfect numbers remains open.

In case you're interested in a different style proof:

The sum of divisors function, $\sigma(n)$ adjusted to not add $n$ itself, gives for

$$n^2=p_1^{2e_1}\ldots p_k^{2e_k}$$

the value

$$\sigma'(n^2)=\prod_{i=1}^k\sigma'(p_i^{2e_i})$$

(since $\sigma'$ is still multiplicative)

For a prime power, $p^j$ we know

$$\sigma'(p^j)={p^{j}-1\over p-1}$$

Note for $j=2m$ this gives divisibility by $p+1$, since $2|m\implies (x^2-1)|(x^{2m}-1)$, hence $(p^2-1)|(p^{2m}-1)$. But since $n$ is odd, $p+1$ must be an even number, since $p$ must be odd. So $2|\sigma'(n^2)$, so $n^2$ is imperfect.