Math Problem Statement

Suppose for integers a, b, and c that a^2 + b^2 = c^2. Then either a or b is divisible by 3.

Solution

This problem is related to properties of integers, divisibility, and the structure of Pythagorean triples. Let's analyze and prove the statement:

Problem:

Given integers aa, bb, and cc satisfying a2+b2=c2a^2 + b^2 = c^2, prove that either aa or bb is divisible by 3.


Proof:

  1. Modulo 3 Analysis:
    Any integer nn can be in one of three residue classes modulo 3: 00, 11, or 22. Specifically: n20,1(mod3).n^2 \equiv 0, 1 \pmod{3}. This is because:

    • If n0(mod3)n \equiv 0 \pmod{3}, then n202=0(mod3)n^2 \equiv 0^2 = 0 \pmod{3}.
    • If n1(mod3)n \equiv 1 \pmod{3}, then n212=1(mod3)n^2 \equiv 1^2 = 1 \pmod{3}.
    • If n2(mod3)n \equiv 2 \pmod{3}, then n222=41(mod3)n^2 \equiv 2^2 = 4 \equiv 1 \pmod{3}.

    Thus, the only possible values for n2mod3n^2 \mod 3 are 00 or 11.

  2. Sum Modulo 3:
    From the equation a2+b2=c2a^2 + b^2 = c^2, consider this modulo 3: a2+b2c2(mod3).a^2 + b^2 \equiv c^2 \pmod{3}. Using the result above, a2a^2, b2b^2, and c2c^2 can each be either 00 or 1(mod3)1 \pmod{3}. We now analyze all possible cases:

    • Case 1: Both a2a^2 and b2b^2 are 1(mod3)1 \pmod{3}.
      Then: a2+b21+1=2(mod3).a^2 + b^2 \equiv 1 + 1 = 2 \pmod{3}. But c2c^2 must be 00 or 1(mod3)1 \pmod{3}, so this case is impossible.

    • Case 2: One of a2a^2 or b2b^2 is 0(mod3)0 \pmod{3}, and the other is 1(mod3)1 \pmod{3}.
      Then: a2+b20+1=1(mod3),a^2 + b^2 \equiv 0 + 1 = 1 \pmod{3}, which matches c2(mod3)c^2 \pmod{3}.

    • Case 3: Both a2a^2 and b2b^2 are 0(mod3)0 \pmod{3}.
      Then: a2+b20+0=0(mod3),a^2 + b^2 \equiv 0 + 0 = 0 \pmod{3}, which matches c2(mod3)c^2 \pmod{3}.

  3. Conclusion:
    For a2+b2=c2a^2 + b^2 = c^2 to hold, at least one of a2a^2 or b2b^2 must be 0(mod3)0 \pmod{3}. This implies that at least one of aa or bb is divisible by 3.


Final Statement:

Thus, it is proven that either aa or bb is divisible by 3.


Do you want further clarification or a more detailed breakdown?
Here are 5 related questions to consider:

  1. What are Pythagorean triples, and how do they relate to this problem?
  2. Can you generalize the proof for other divisors, such as 5 or 7?
  3. How does the concept of modular arithmetic simplify proofs in number theory?
  4. What are the properties of numbers modulo nn for different nn?
  5. Can this problem be approached geometrically or visually?

Tip: Understanding modular arithmetic is crucial for number theory proofs. Practice analyzing equations modulo small integers!

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Math Problem Analysis

Mathematical Concepts

Number Theory
Divisibility
Pythagorean Triples
Modular Arithmetic

Formulas

a^2 + b^2 = c^2

Theorems

Pythagorean Theorem
Modular Arithmetic Properties

Suitable Grade Level

Grades 10-12