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Archangel Macsika

Let x be an integer. Prove that if x^2 − 6x + 5 is even then x must be odd.

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Solution

Proof.

Assume for the sake of contradiction that x2 − 6x + 5 is even and x is even.

By definition of even, x = 2c where c is an integer.

By substitution,

x2 − 6x + 5 = (2c)2 − 6(2c) + 5

= 4c2 − 12c + 5

= 4c2 − 12c + 4 + 1

= 2(2c2 − 6c + 2) + 1.

Since 2c2 − 6c + 2 is an integer, x2 − 6x + 5 must be odd, which is a contradiction.



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