SOLUTION: Determine the truth value of the following statement.
If n = 4k + 1 for a natural k, then there are an even natural number a and an odd natural number b
such that n = a^2+b^2
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-> SOLUTION: Determine the truth value of the following statement.
If n = 4k + 1 for a natural k, then there are an even natural number a and an odd natural number b
such that n = a^2+b^2
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Question 1019875: Determine the truth value of the following statement.
If n = 4k + 1 for a natural k, then there are an even natural number a and an odd natural number b
such that n = a^2+b^2
Write the converse of this statement and determine its truth value as well. Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! The statement is not always true. Take for example k = 5. Then n = 4*5 + 1 = 21, but 21 is not the sum of squares of odd and even natural numbers. (Another is when k = 8.)
The converse goes this way:
For all even natural number a and odd natural number b such that n = a^2+b^2, then there is a natural number k such that n = 4k + 1. (And because a statement and its converse are equivalent, the converse is also not true.)
