Prime Divisibility Theorem Explained with Proof

 

❓ Theorem

Let $p$ be a prime number.

If $p$ divides $a^2$, then $p$ also divides $a$, where $a$ is a positive integer.

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🖼️ Solution Image

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✍️ Short Explanation

This theorem is based on the Fundamental Theorem of Arithmetic.

If a prime number is a factor of $a^2$, then it must already be present in the prime factorisation of $a$ 💯

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🔹 Step 1 — Write Prime Factorisation of $a$

Let:

$$a = p_1 \times p_2 \times p_3 \times \cdots \times p_n$$

where:

$$p_1, p_2, p_3, \ldots$$

are prime factors.

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🔹 Step 2 — Square Both Sides

$$a^2 = p_1^2 \times p_2^2 \times p_3^2 \times \cdots \times p_n^2$$

👉 Every prime factor of $a$ appears twice in $a^2$.

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🔹 Step 3 — Apply Given Condition

Given:

$$p \mid a^2$$

This means $p$ divides $a^2$.

By the Fundamental Theorem of Arithmetic, prime factorisation is unique.

So $p$ must be one of the prime factors of:

$$a^2$$

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🔹 Step 4 — Final Conclusion

Since $p$ is a prime factor of $a^2$, it must also be present in the factorisation of:

$$a$$

Hence,

$$p \mid a$$

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✅ Final Statement

$$\boxed{\text{If } p \mid a^2,\ \text{then } p \mid a}$$

where $p$ is a prime number.

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⭐ Key Insight

• Prime factors of $a$ repeat in $a^2$
  
• If a prime divides the square, it must divide the original number too

🧠 Memory Line:

Prime dividing a square always divides the number itself