Question 579097
Intuitively, we can guess it is 7 and 9, which works, but to solve it mathematically

2n+1, 2n+3

{{{sqrt(2n+3) = 2n+1-4}}}
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{{{sqrt(2n+3) = 2n-3}}}
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Square both sides:
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2n+3 = 4n^2 -12n + 9

4n^2 - 14n + 6 = 0;   divide by 2
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2n^2 - 7n + 3 = 0; factor:
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(2n - 1)(n - 3) = 0

n = 1/2 or 3

plugging 1/2 into original equations gives us 2 and 4 which are even, so throw that out

plugging 3 into original equations gives us 7 and 9, which work

*[invoke quadratic "x", 2, -7, 3 ]