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Suppose aa, bb and cc are integers. If a2+b2=c2a^2 + b^2 = c^2 (i.e., a2+b2a^2 + b^2 is a perfect square), show that aa or bb is even.

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Q. Suppose aa, bb and cc are integers. If a2+b2=c2a^2 + b^2 = c^2 (i.e., a2+b2a^2 + b^2 is a perfect square), show that aa or bb is even.
  1. Assume odd integers: Assume both aa and bb are odd integers.
  2. Express aa and bb: Let a=2k+1a = 2k+1 and b=2m+1b = 2m+1, where kk and mm are integers.
  3. Calculate a2a^2: Calculate a2a^2: (2k+1)2=4k2+4k+1(2k+1)^2 = 4k^2 + 4k + 1.
  4. Calculate b2b^2: Calculate b2b^2: (2m+1)2=4m2+4m+1(2m+1)^2 = 4m^2 + 4m + 1.
  5. Add a2a^2 and b2b^2: Add a2a^2 and b2b^2: 4k2+4k+1+4m2+4m+14k^2 + 4k + 1 + 4m^2 + 4m + 1.
  6. Simplify the sum: Simplify the sum: 4k2+4k+4m2+4m+24k^2 + 4k + 4m^2 + 4m + 2.
  7. Factor out common factor: Factor out the common factor of 44: 4(k2+k+m2+m)+24(k^2 + k + m^2 + m) + 2.
  8. Identify contradiction: Notice that the sum is 22 more than a multiple of 44, which cannot be a perfect square since all perfect squares are either 00, 11, or 44 modulo 44.
  9. Conclusion: Since our assumption leads to a contradiction, at least one of aa or bb must be even.

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