discrete math
posted by romulo .
Prove by contradiction that for any even integer a and any odd integer b, 4 does not divide (a^2 + 2b^2).
Proposition: That 4k (k is any integer) = a^2 +2b^2, and a is even, and b is odd.
But 4k is even (product of any integer and 4), so a^2 must be even, as 2b^2 is even.
Dividing both sides by 4,
k=a^2/4 + 2b^2/4
but a is even, so a=2*n where n is an integer. a^2=4n^2
k= n^2 + b^2/2
But b is odd, so b^2/2 is not an integer.
Therefore, k cannot be an integer, so the proposition is contradicted.
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