Prove that for any integer x, the value of x^2+7x+18 is not divisible by 529.
529=23^2
If x^2+7x+18 = k*(23^2)
x^2+7x = k*(23^2)-18
4x^2+28x = 4*k*(23^2)-72
(2x+7)^2 = 4*k*(23^2)-23
Then 23 divides LHS. Since 23 is prime, 23 divides (2x+7) and 23^2 divides LHS. But RHS is never divisible by 23^2 and so the original assumption of even division must be incorrect.
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Posted by xdog
on 2016-07-25 17:47:27 |
completing the square
