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(D) For statement $I$: We check if $n^2+3 \equiv 0 \pmod{17}$.This implies $n^2 \equiv -3 \equiv 14 \pmod{17}$.The quadratic residues modulo $17$ are

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$I$ is true and $II$ is false.

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(D) For statement $I$: We check if $n^2+3 \equiv 0 \pmod{17}$.This implies $n^2 \equiv -3 \equiv 14 \pmod{17}$.The quadratic residues modulo $17$ are $0^2=0, 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25 \equiv 8, 6^2=36 \equiv 2, 7^2=49 \equiv 15, 8^2=64 \equiv 13$.Since $14$ is not in the set of quadratic residues ${0, 1, 2, 4, 8, 9, 13, 15, 16}$,$n^2+3$ is never divisible by $17$. Thus,$I$ is true.For statement $II$: We check if $n^2+4 \equiv 0 \pmod{17}$.This implies $n^2 \equiv -4 \equiv 13 \pmod{17}$.From the list of quadratic residues above,$8^2 = 64 = 3 	imes 17 + 13$,so $8^2 \equiv 13 \pmod{17}$.For $n=8$,$n^2+4 = 64+4 = 68 = 4 	imes 17$,which is divisible by $17$.Thus,$II$ is false.Therefore,$I$ is true and $II$ is false.

Consider the following statements: For any integer $n$,
$I.$ $n^2+3$ is never divisible by $17$.
$II.$ $n^2+4$ is never divisible by $17$.
Then,

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Consider the following statements: For any integer $n$,$I.$ $n^2+3$ is never divisible by $17$.$II.$ $n^2+4$ is never divisible by $17$.Then,