Is There a Name for This Theorem?

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Discussion Overview

The discussion revolves around the existence of a theorem related to the divisibility condition for integers, specifically examining the implication that if \( b \) divides \( a^2 \), then \( b \) must also divide \( a \). Participants explore the conditions under which this implication holds and seek to identify any formal name for such a theorem.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions whether there is a theorem stating that if \( b|a^2 \), then \( b|a \) is true, suggesting it is not generally true without certain conditions.
  • Another participant provides a counterexample, noting that while \( 9|36 \) is true, \( 9|6 \) is not, indicating that \( b \) must be prime or have no repeated prime factors for the implication to hold.
  • A participant clarifies that the implication \( b|a \rightarrow b|a^2 \) is indeed true, but suggests it may not warrant being called a theorem as it is a special case of a more general divisibility rule.
  • One participant asserts that the implication holds true whenever \( |\mu(b)|=1 \), hinting at a specific mathematical condition related to the divisor \( b \).
  • Another participant references Euclid's lemma to support the claim that if \( b \) is prime and \( b|a^2 \), then \( b|a \) must also be true.

Areas of Agreement / Disagreement

Participants express disagreement regarding the generality of the theorem in question, with some asserting it holds under specific conditions (e.g., when \( b \) is prime) while others argue it is not universally applicable.

Contextual Notes

The discussion highlights the need for specific conditions for the divisibility implication to hold, such as the primality of \( b \) or the absence of repeated prime factors. There is also mention of a mathematical condition involving \( |\mu(b)| \) that remains unexplained.

Dschumanji
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Is there a theorem that says when b|a2 → b|a is true for integers a and b?

If so, what is it called?
 
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I hope not, since it isn't generally true. 9|36, but not 9|6. You would need that b is a prime (or at least, has no repeated prime factors).
 
OP, did you mean to reverse those...?

b|a \; \rightarrow \; b|a^{2}

Is certainly true.
 
Last edited:
daniel.e2718 said:
Is certainly true.

But it's hardly worth calling it a theorem, since it's just a special case of ##b|a \rightarrow b|ac##.
 
Dschumanji said:
Is there a theorem that says when b|a2 → b|a is true for integers a and b?

If so, what is it called?

That is true whenever b is prime. You can prove it by using euclid's lemma.

Let b be prime. Suppose b|a2. Then b|aa, and, by euclid's lemma, b|a or b|a. Hence b|a.
 
The statement holds true whenever |\mu(b)|=1.
 

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