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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- Jun 16th 2014, 10:49 PMBenja303Indirect Proof question
Using and indirect proof prove that if n is an integer and 5n^2+ 19 is even, then n is

odd. - Jun 16th 2014, 11:22 PMromsekRe: Indirect Proof question
- Jun 21st 2014, 04:29 AMHallsofIvyRe: Indirect Proof question
A little more detail: Suppose n were even. Then n= 2k for some integer k. $\displaystyle 5n^2+ 19= 5(2k)^2+ 19= 5(4k^2)+ 19= 20k^2+ 19= 20k^2+ 18+ 1= 2(10k^2+ 9)+ 1$. Since $\displaystyle 5n^2+ 19$ is equal to 2 times an integer plus one, it is odd, not even, a contradiction.