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mathisfun1 · #2 August 6th, 2007, 12:55 PM

The function is symmetric about x=2 and x=7, so all necessary zero are found by reflecting x=0 repeatedly about x=2 and x=7; if $R_2$ is a reflection about x=2 and $R_7$ about x=7, then we obtain a sequence of zeros $R_2, \ R_7R_2, \ R_2R_7R_2, ...$ or $R_7, \ R_2R_7, \ R_7R_2R_7, ...$ . If $R_2$ is our first reflection, then we obtain the following sequence of zeros $a_0=0, \ a_1=4, \ a_2 = 10, \ a_3 = -6, \ a_4 = 20$ and in general $a_{2m}=10m, \ a_{2m-1} = 14-10m$ , giving us 200+202 = 402 zeros in the interval [-2007, 2007]. If $R_7$ is our first reflection, then we get the sequence $a_{2k}=-10k, \ a_{2k-1} = 10k+4$ , which yields another 402 zeros. A quick check shows that the two sequences have no terms in common. Hence at least 804 zeros total in [-2007, 2007].