Prime numbers proof by contradiction

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SUMMARY

The discussion focuses on proving that for prime numbers \(a\), \(b\), and \(c\), the equation \(a^2 + b^2 \ne c^2\) holds true through contradiction. Participants analyze the equation \(a^2 = c^2 - b^2 = (c - b)(c + b)\) and explore the implications of prime number properties. The proof requires understanding the relationships between prime numbers and their factors, ultimately leading to the conclusion that the initial assumption of equality leads to contradictions in the properties of primes.

PREREQUISITES
  • Understanding of prime numbers and their properties
  • Familiarity with mathematical proof techniques, particularly proof by contradiction
  • Basic algebraic manipulation and factorization
  • Knowledge of number theory concepts related to prime factorization
NEXT STEPS
  • Study the principles of proof by contradiction in mathematics
  • Explore the properties of prime numbers in number theory
  • Learn about algebraic identities and their applications in proofs
  • Investigate the implications of the Pythagorean theorem in relation to prime numbers
USEFUL FOR

Mathematicians, students studying number theory, educators teaching proof techniques, and anyone interested in the properties of prime numbers and mathematical reasoning.

tmt1
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For prime numbers, $a$, $b$, $c$, $a^2 + b^2 \ne c^2$. Prove this by contradiction.

So, I get that $a^2 = c^2 - b^2 = (c - b)(c +b)$

And I get that prime numbers are the product of 2 numbers that are either greater than one, or less than the prime numbers.

But I'm unsure how to go from here.
 
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tmt said:
For prime numbers, $a$, $b$, $c$, $a^2 + b^2 \ne c^2$. Prove this by contradiction.

So, I get that $a^2 = c^2 - b^2 = (c - b)(c +b)$

And I get that prime numbers are the product of 2 numbers that are either greater than one, or less than the prime numbers.

But I'm unsure how to go from here.

a is prime so either $c-b = 1$ and $c+b= a^2$ or $c-b=c+b=a$ can you proceed from here
 

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