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Indivisibility by 529 (Posted on 2016-07-25) Difficulty: 3 of 5
Prove that for any integer x, the value of x^2+7x+18 is not divisible by 529.

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completing the square | Comment 2 of 3 |

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 2016-07-25 17:47:27
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