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.

Posted by xdog
on 20160725 17:47:27 