If a and are any odd integers, then a^{2} + b^{2} is even. Prove?

djeljenike 2020-10-27 Answered
If a and are any odd integers, then a2 + b2 is even.
Prove?
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Expert Answer

l1koV
Answered 2020-10-28 Author has 100 answers

Proof:
Let two odd positive numbers be a=2j + 1 and b=2k + 1: where j and k are integers.
a2 + b2=(2j + 1)2 + (2k + 1)2
=4j2 + 4j + 1 + 4k2 + 4k + 1
=2(2j2 + 2j) + 1 + 2(2k2 + 2k) + 1
=2l + 2m + 2
where, l=2j2 + 2j, m=2k2 + 2k are integers. Then,
a2 + b2=2(l + m + 1)
a2 + b2=2n which is even by definition.
where, n=l + m + 1 is also an integer.
Hence, if a and b are any odd integers then a2 + b2 is even.
Conclusion:
The statement, "If a and b are any odd integers, then a2 + b2 is even" is proved.

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Jeffrey Jordon
Answered 2021-11-03 Author has 2047 answers

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