Kvant Math Problem 223
Perfect numbers are rare and highly structured.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m03s
Source on kvant.digital
Problem
A natural number is called perfect if it is equal to the sum of all its divisors except the number itself. (For example, 28 is a perfect number: $28=1+2+4+7+14$.) Prove that a perfect number cannot be a perfect square.
A. Makarichev, 9th-grade student
Exploration
Perfect numbers are rare and highly structured. The first few examples are $6$, $28$, $496$, and $8128$. Checking small perfect numbers, none are perfect squares: $6=2\cdot3$, $28=2^2\cdot7$, $496=2^4\cdot31$, $8128=2^6\cdot127$. Their prime factorizations suggest a connection to powers of two multiplied by a Mersenne prime. Considering whether a perfect number could be a square, one might attempt small exponents in these factorizations. For example, $28=2^2\cdot7$; $2^2\cdot7$ is not a perfect square. If the general even perfect number has the form $2^{p-1}(2^p-1)$ with $2^p-1$ prime, we can attempt to see whether this can ever be a square. Since a square has all even exponents in its prime factorization, and $2^p-1$ is odd and prime, $2^p-1$ would need an even exponent, which is impossible. This suggests the crucial point is the structure of even perfect numbers. Odd perfect numbers are hypothetical; even if they exist, similar parity arguments on the sum-of-divisors function will likely yield a contradiction. The most likely subtlety is ensuring the prime factorization argument applies in all cases and not just the small examples.
Problem Understanding
The problem asks to prove that no perfect number is a perfect square. This is a Type B problem. The core difficulty lies in connecting the definition of a perfect number, which is additive in terms of divisors, to multiplicative properties captured by prime factorization, particularly the exponents in a square. The central insight is that the canonical form of perfect numbers forces an odd prime factor to appear to the first power, which is incompatible with being a perfect square.
Proof Architecture
Lemma 1: Every even perfect number has the form $2^{p-1}(2^p-1)$ where $2^p-1$ is prime. This follows from the Euclid–Euler theorem.
Lemma 2: If $2^{p-1}(2^p-1)$ were a perfect square, then both $2^{p-1}$ and $2^p-1$ would have to be squares. This uses the uniqueness of prime factorizations.
Lemma 3: The prime $2^p-1$ cannot be a perfect square for $p>2$. If $2^p-1 = k^2$, then $k^2+1$ is a power of two; solving $k^2+1=2^p$ only allows $k=1$ and $p=1$, which is too small.
Lemma 4: Odd perfect numbers, if they exist, have the form $N = q^{\alpha} m^2$ with $q$ prime and $\alpha \equiv 1 \pmod 4$. Therefore $N$ cannot be a perfect square because the exponent of $q$ is odd.
The hardest part is rigorously excluding even perfect numbers being squares using the form $2^{p-1}(2^p-1)$, and the subtlety lies in solving $k^2+1=2^p$. The lemma most likely to fail is the odd perfect number argument, as the existence of such numbers is not known.
Solution
Let $n$ be a perfect number. First, assume $n$ is even. By the Euclid–Euler theorem, any even perfect number has the form $n = 2^{p-1}(2^p-1)$, where $2^p-1$ is prime. Assume for contradiction that $n$ is a perfect square. Then $2^{p-1}(2^p-1) = k^2$ for some integer $k$. Consider the prime factorizations. In $2^{p-1}(2^p-1)$, the exponent of $2$ is $p-1$, so $p-1$ must be even. Hence $p$ is odd, so $p \ge 3$. Then $2^p-1$ is an odd prime. Since $2^p-1$ is prime and appears with exponent $1$, for $k^2$ to have all even exponents in its prime factorization, $2^p-1$ would also need an even exponent. This is impossible, since its exponent is $1$. Therefore $n$ cannot be an even perfect square.
Next, consider an odd perfect number. Any odd perfect number, if it exists, can be written in the form $N = q^{\alpha} m^2$, where $q$ is prime, $q \equiv \alpha \equiv 1 \pmod 4$, and $\gcd(q,m)=1$. The exponent of $q$ is $\alpha$, which is odd. Since a perfect square must have all exponents even in its prime factorization, $N$ cannot be a perfect square.
Thus, no perfect number, whether even or odd, can be a perfect square. This completes the proof.
∎
Verification of Key Steps
The step excluding even perfect numbers as squares relies on solving $2^{p-1}(2^p-1)=k^2$. Checking small cases: $p=2$ gives $n=2^{1}(3)=6$, not a square; $p=3$ gives $2^{2}\cdot7=28$, not a square; $p=5$ gives $2^{4}\cdot31=496$, not a square. The equation $2^p-1=k^2$ reduces to $k^2+1=2^p$, whose only positive integer solution is $k=1, p=1$, which does not produce a perfect number. Therefore, the general argument is validated.
For odd perfect numbers, the exponent $\alpha$ of the prime $q$ being odd ensures that $q^{\alpha}$ cannot be part of a square, which is an entirely arithmetic argument independent of the unknown existence of odd perfect numbers. Any careless argument that ignored $\alpha$’s parity would fail.
Alternative Approaches
An alternative approach for even perfect numbers could use modular arithmetic: any perfect square is congruent to $0$ or $1 \pmod 4$, whereas $2^{p-1}(2^p-1)$ with $p\ge3$ is congruent to $6 \pmod 8$, which cannot be a square. This avoids explicit factorization considerations but is less general because it does not extend to hypothetical odd perfect numbers. The chosen approach is preferable because it unifies both even and odd perfect numbers and relies solely on prime factorization and exponents, providing a fully rigorous, general argument.