Using and indirect proof prove that if n is an integer and 5n^2+ 19 is even, then n is odd.
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Originally Posted by Benja303 Using and indirect proof prove that if n is an integer and 5n^2+ 19 is even, then n is odd. Suppose n were even. $5 n^2 + 19=5(2k)^2+19=20k^2+19$ $20 k^2 + 19$ is clearly odd for any k and thus by contradiction n must be odd.
A little more detail: Suppose n were even. Then n= 2k for some integer k. . Since is equal to 2 times an integer plus one, it is odd, not even, a contradiction.
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