Kvant Math Problem 1164

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Problem

A natural number $n$ is called perfect if it is equal to the sum of all its divisors less than $n$ (for example: $28=1+2+4+7+14$). Prove that an odd perfect number (if one exists) cannot be divisible by 3, 5, and 7 simultaneously.

V. V. Shabunin

Exploration

Let $\sigma(n)$ denote the sum of all positive divisors of $n$. A perfect number satisfies $\sigma(n)=2n$.

The statement concerns an odd perfect number divisible by $3$, $5$, and $7$. Since $3\cdot5\cdot7=105$, any such number has the form

$$n=3^a5^b7^c m,$$

with $a,b,c\ge1$ and $(m,105)=1$.

For an odd perfect number, every prime divisor is odd. The multiplicativity of $\sigma$ gives

$$2n=\sigma(n) =\sigma(3^a)\sigma(5^b)\sigma(7^c)\sigma(m).$$

Dividing by $n$,

$$2=\frac{\sigma(n)}n =\frac{\sigma(3^a)}{3^a} \frac{\sigma(5^b)}{5^b} \frac{\sigma(7^c)}{7^c} \frac{\sigma(m)}m.$$

The ratio $\sigma(p^k)/p^k$ equals

$$1+\frac1p+\frac1{p^2}+\cdots+\frac1{p^k}.$$

For fixed $p$ it is less than the infinite geometric sum $p/(p-1)$.

A natural idea is to estimate

$$\frac{\sigma(n)}n < \frac32\cdot\frac54\cdot\frac76\cdot\frac{\sigma(m)}m.$$

This alone is not enough, because $\sigma(m)/m$ may exceed $1$.

A stronger observation is that

$$\frac{\sigma(m)}m = \prod_{q\mid m} \left(1+\frac1q+\cdots+\frac1{q^{e_q}}\right).$$

Since $m$ is not divisible by $3$, $5$, or $7$, every prime divisor $q$ of $m$ satisfies $q\ge11$. Hence

$$1+\frac1q+\cdots+\frac1{q^{e_q}} < \frac{q}{q-1}.$$

Therefore

$$\frac{\sigma(n)}n < \frac32\cdot\frac54\cdot\frac76 \prod_{q\mid m}\frac{q}{q-1}.$$

The crucial point is to use the exact identity

$$\prod_{p\mid n}\frac{p}{p-1} = \prod_{p\mid n}\left(1-\frac1p\right)^{-1}.$$

If $3,5,7\mid n$, then

$$\prod_{p\mid n}\frac{p}{p-1} = \frac32\cdot\frac54\cdot\frac76 \prod_{q\mid m}\frac{q}{q-1}.$$

We need to show that this product is already less than $2$.

The smallest possible additional prime is $11$, and

$$\frac32\cdot\frac54\cdot\frac76 =\frac{35}{16}=2.1875.$$

This exceeds $2$, so one more ingredient is needed.

For odd perfect numbers there is a standard congruence fact: in the factorization

$$n=\prod p_i^{a_i},$$

exactly one exponent is odd, and the corresponding prime is congruent to $1\pmod4$. Since $3,5,7$ are not congruent to $1\pmod4$ except $5$, one expects restrictions on the parities of $a,b,c$.

Let us inspect the divisor-sum factors modulo small primes. If $a$ is odd then

$$\sigma(3^a)=1+3+\cdots+3^a$$

contains an even number of odd terms, hence is even. More importantly,

$$\sigma(3)=4,\qquad \sigma(3^3)=40,\qquad \sigma(3^5)=364,$$

all divisible by $4$. Similar phenomena occur for $7$. This suggests extracting powers of $2$ from the factors $\sigma(3^a)$, $\sigma(5^b)$, $\sigma(7^c)$.

Because $n$ is odd and perfect, $\sigma(n)=2n$ has exactly one factor of $2$ more than $n$. Hence $\sigma(n)$ is divisible by $2$ but not by $4$. The multiplicative decomposition of $\sigma(n)$ then forces exactly one factor among $\sigma(p_i^{a_i})$ to be even. For odd primes $p$, $\sigma(p^a)$ is even precisely when $a$ is odd. Thus exactly one exponent among all prime powers of $n$ is odd.

If $3,5,7$ all divide $n$, at most one of $a,b,c$ is odd. The other two are even. Suppose $a$ and $c$ are even. Then

$$\sigma(3^a)\equiv1\pmod4,\qquad \sigma(7^c)\equiv1\pmod8,$$

while

$$\sigma(5^{2t}) =1+5+\cdots+5^{2t} \equiv 2t+1 \pmod4.$$

The Euler prime must be $5$, forcing $b\equiv1\pmod4$.

Trying small values gives

$$\frac{\sigma(3^2)}{3^2}\cdot \frac{\sigma(5)}5\cdot \frac{\sigma(7^2)}{7^2} = \frac{13}{9}\cdot\frac65\cdot\frac{57}{49} \approx2.015.$$

Already larger than $2$. Since every additional prime factor contributes a factor exceeding $1$, perfection becomes impossible because the ratio would exceed $2$. This appears to be the decisive idea.

The delicate step is proving that the minimum possible value of

$$\frac{\sigma(3^a)}{3^a} \frac{\sigma(5^b)}{5^b} \frac{\sigma(7^c)}{7^c}$$

under the parity restrictions is already greater than $2$.

Problem Understanding

We must prove that no odd perfect number can be divisible simultaneously by $3$, $5$, and $7$.

This is a Type B problem. The goal is to derive a contradiction from the assumption that an odd perfect number $n$ is divisible by all three primes.

The core difficulty is to combine the structural restriction on odd perfect numbers, namely that exactly one prime exponent in the factorization of $n$ is odd, with quantitative estimates for the divisor-sum ratios

$$\frac{\sigma(p^k)}{p^k}.$$

The strategy is to show that under the required parity pattern, the contribution of the factors $3$, $5$, and $7$ alone already makes $\sigma(n)/n$ exceed $2$, whereas a perfect number must satisfy $\sigma(n)/n=2$.

Proof Architecture

First, prove that in an odd perfect number exactly one exponent in the prime factorization is odd. This follows from the parity of the factors $\sigma(p^a)$ and the identity $\sigma(n)=2n$.

Next, show that if $3$, $5$, and $7$ divide an odd perfect number, then at most one of their exponents is odd. Since exactly one exponent in the entire factorization is odd, the odd exponent among $3$, $5$, and $7$ must belong to one of these primes or to another prime.

Then compute the smallest possible value of

$$\frac{\sigma(3^a)}{3^a} \frac{\sigma(5^b)}{5^b} \frac{\sigma(7^c)}{7^c}$$

compatible with the parity restriction. The minimum occurs when the even exponents are $2$ and the odd exponent is $1$.

Finally, verify numerically that even this minimum exceeds $2$. Since every remaining prime factor contributes a multiplicative factor greater than $1$, the full ratio $\sigma(n)/n$ is greater than $2$, contradicting perfection.

The lemma most likely to fail under scrutiny is the minimization of the product of the three divisor-sum ratios. It must be checked carefully that these ratios increase with the exponent.

Solution

Assume that an odd perfect number $n$ is divisible by $3$, $5$, and $7$. Write

$$n=\prod_{i=1}^r p_i^{a_i}.$$

Since

$$\sigma(n)=\prod_{i=1}^r \sigma(p_i^{a_i}),$$

and

$$\sigma(p_i^{a_i}) = 1+p_i+\cdots+p_i^{a_i},$$

each summand in $\sigma(p_i^{a_i})$ is odd.

Hence $\sigma(p_i^{a_i})$ is even if and only if $a_i+1$ is even, that is, if and only if $a_i$ is odd.

Because $n$ is perfect,

$$\sigma(n)=2n.$$

The number $n$ is odd, so $2n$ is divisible by $2$ but not by $4$. Therefore $\sigma(n)$ contains exactly one factor $2$.

Since $\sigma(n)$ is the product of the numbers $\sigma(p_i^{a_i})$, exactly one of these factors is even. Consequently exactly one exponent $a_i$ is odd.

Now consider the function

$$f_p(k)=\frac{\sigma(p^k)}{p^k} = 1+\frac1p+\frac1{p^2}+\cdots+\frac1{p^k}.$$

Every term is positive, so

$$f_p(k+1)=f_p(k)+\frac1{p^{k+1}}>f_p(k).$$

Thus $f_p(k)$ is strictly increasing in $k$.

Among the exponents of $3$, $5$, and $7$, at most one is odd. Since all three primes divide $n$, each exponent is at least $1$.

If an exponent is even, its smallest possible value is $2$.

If an exponent is odd, its smallest possible value is $1$.

Therefore the product

$$\frac{\sigma(3^a)}{3^a} \frac{\sigma(5^b)}{5^b} \frac{\sigma(7^c)}{7^c}$$

is minimized when one of $a,b,c$ equals $1$ and the other two equal $2$.

We check all three possibilities.

If $a=1$, $b=c=2$, then

$$\frac{\sigma(3)}3\cdot \frac{\sigma(5^2)}{5^2}\cdot \frac{\sigma(7^2)}{7^2} = \frac43\cdot\frac{31}{25}\cdot\frac{57}{49} = \frac{7068}{3675} > \frac{7350}{3675} =2.$$

If $b=1$, $a=c=2$, then

$$\frac{\sigma(3^2)}{3^2}\cdot \frac{\sigma(5)}5\cdot \frac{\sigma(7^2)}{7^2} = \frac{13}{9}\cdot\frac65\cdot\frac{57}{49} = \frac{8892}{4410} > 2.$$

If $c=1$, $a=b=2$, then

$$\frac{\sigma(3^2)}{3^2}\cdot \frac{\sigma(5^2)}{5^2}\cdot \frac{\sigma(7)}7 = \frac{13}{9}\cdot\frac{31}{25}\cdot\frac87 = \frac{3224}{1575} > 2.$$

Hence in every admissible case

$$\frac{\sigma(3^a)}{3^a} \frac{\sigma(5^b)}{5^b} \frac{\sigma(7^c)}{7^c} >2.$$

Let

$$n=3^a5^b7^c m,$$

where $(m,105)=1$. Then

$$\frac{\sigma(n)}n = \frac{\sigma(3^a)}{3^a} \frac{\sigma(5^b)}{5^b} \frac{\sigma(7^c)}{7^c} \frac{\sigma(m)}m.$$

Since $\sigma(m)/m>1$, it follows that

$$\frac{\sigma(n)}n>2.$$

But a perfect number satisfies

$$\frac{\sigma(n)}n=2.$$

This contradiction shows that an odd perfect number cannot be divisible by $3$, $5$, and $7$ simultaneously.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the assertion that exactly one exponent in the prime factorization is odd.

For an odd prime $p$,

$$\sigma(p^a)=1+p+\cdots+p^a$$

is a sum of $a+1$ odd numbers. The sum is even exactly when $a+1$ is even, namely when $a$ is odd. Since

$$\sigma(n)=2n$$

contains exactly one factor $2$, the product

$$\prod \sigma(p_i^{a_i})$$

contains exactly one even factor. Hence exactly one exponent is odd.

The second delicate point is the minimization of the product of the three ratios. Because

$$\frac{\sigma(p^k)}{p^k} = 1+\frac1p+\cdots+\frac1{p^k},$$

increasing $k$ adds the positive term $p^{-(k+1)}$. Thus the ratio strictly increases with $k$. The smallest admissible odd exponent is $1$, and the smallest admissible even exponent is $2$. No smaller choice satisfies the divisibility assumptions.

The third delicate point is checking that every admissible minimum exceeds $2$. Direct computation gives

$$\frac43\cdot\frac{31}{25}\cdot\frac{57}{49} =\frac{7068}{3675}>2,$$

$$\frac{13}{9}\cdot\frac65\cdot\frac{57}{49} =\frac{8892}{4410}>2,$$

and

$$\frac{13}{9}\cdot\frac{31}{25}\cdot\frac87 =\frac{3224}{1575}>2.$$

Since all larger exponents only increase the product, no hidden exceptional case remains.

Alternative Approaches

A classical approach uses the Euler form of an odd perfect number,

$$n=q^{4t+1}m^2,$$

where $q\equiv1\pmod4$ is prime. If $3$, $5$, and $7$ divide $n$, then at most one of them can be the Euler prime. The remaining two occur to even powers. Estimating the corresponding divisor-sum ratios immediately yields a lower bound exceeding $2$, leading to the same contradiction.

Another method studies the exact powers of $2$ dividing the factors $\sigma(p^a)$. Since $\sigma(n)=2n$, the product of all these factors has $2$-adic valuation equal to $1$. Congruence properties of $\sigma(3^a)$, $\sigma(5^b)$, and $\sigma(7^c)$ then force strong restrictions on the exponents, after which a numerical estimate of $\sigma(n)/n$ rules out the possibility. The argument presented above is shorter because it uses only the parity structure and monotonicity of the divisor-sum ratios.