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February 1, 2015

February 1, 2015

Posted by **romulo** on Wednesday, March 7, 2007 at 1:45am.

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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