Proof that all and only perfect squares have an odd number of different factors

Show that a natural number has an odd number of different factors if, and only if, it is a perfect square.

Proof: A natural number has an odd number of different factors if and only if it is a perfect square.

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Step 1: Prime Factorization Approach

Let’s express a natural number $n$ in terms of its prime factorization:

$$ n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k} $$

where $p_1, p_2, \dots, p_k$ are distinct prime numbers and $a_1, a_2, \dots, a_k$ are non-negative integers.

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Step 2: Total Number of Factors Formula

The total number of positive factors of $n$ is given by the formula:

$$ \text{Number of factors} = (a_1 + 1)(a_2 + 1) \dots (a_k + 1) $$

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Step 3: When is the Number of Factors Odd?

The product of numbers $(a_1 + 1)(a_2 + 1) \dots (a_k + 1)$ is odd if and only if each $(a_i + 1)$ is odd, because the product of odd numbers is odd, and if any factor is even, the whole product is even.

A number $(a_i + 1)$ is odd if and only if $a_i$ is even (since an even number plus 1 is odd).

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Step 4: Link to Perfect Squares

• If all exponents $a_i$ are even, then the number is a perfect square, because:

$$ n = (p_1^{a_1/2} p_2^{a_2/2} \dots p_k^{a_k/2})^2 $$

• Conversely, if the number is a perfect square, then all exponents in its prime factorization must be even.

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Step 5: Conclusion

• A natural number has an odd number of positive divisors if and only if it is a perfect square.

✔️ Q.E.D.

Why is the number of factors $(a_1 + 1)(a_2 + 1) \dots (a_k + 1)$?

Excellent question! Let’s break down why the number of factors of a natural number is given by:

$$ (a_1 + 1)(a_2 + 1) \dots (a_k + 1) $$

✅ Reasoning:

Suppose a natural number $n$ has prime factorization:

$$ n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k} $$

🔍 How do we form a factor of $n$?

A factor of $n$ is a number that divides $n$ exactly. To construct a factor, you can choose:

• Any exponent for $p_1$ between $0$ and $a_1$ (inclusive),
• Any exponent for $p_2$ between $0$ and $a_2$ (inclusive),
• ...
• Any exponent for $p_k$ between $0$ and $a_k$ (inclusive).

🔢 Counting the choices:

• For prime $p_1$, there are $(a_1 + 1)$ choices (exponents $0, 1, 2, \dots, a_1$).
• For prime $p_2$, there are $(a_2 + 1)$ choices.
• ...
• For prime $p_k$, there are $(a_k + 1)$ choices.

🔗 Total number of distinct factors:

Each choice of exponent is independent, so the total number of different factors is the product:

$$ (a_1 + 1)(a_2 + 1) \dots (a_k + 1) $$

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📜 Example to illustrate:

Let’s take $n = 12$.

• Prime factorization: $12 = 2^2 \times 3^1$.
• Number of factors:

$$ (2+1)(1+1) = 3 \times 2 = 6 $$

• The factors are: $1, 2, 3, 4, 6, 12$ — 6 factors.

✔️ The formula holds perfectly.

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🚩 Conclusion:

The formula works because forming a divisor is about independently choosing each exponent from its possible range $0$ to $a_i$ for every prime factor.

Would you like me to visualize this with a tree diagram or table?