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Number theory Difficulty 6.0 National olympiad Prove it

7. Prove that for any integer nn, we have
(i) 6n(n+1)(n+2)6 \mid n(n+1)(n+2);
(ii) 8n(n+1)(n+2)(n+3)8 \mid n(n+1)(n+2)(n+3);
(iii) 24n(n+1)(n+2)(n+3)24 \mid n(n+1)(n+2)(n+3);
(iv) if 2n2 \nmid n, then 8n218 \mid n^{2}-1 and 24n(n21)24 \mid n\left(n^{2}-1\right);
(v) if 2n2 \nmid n and 3n3 \nmid n, then 24n2+2324 \mid n^{2}+23;
(vi) 6n3n6 \mid n^{3}-n;
(vii) 30n5n30 \mid n^{5}-n;
(viii) 42n7n42 \mid n^{7}-n;
(ix) 15n5+13n3+715n\frac{1}{5} n^{5}+\frac{1}{3} n^{3}+\frac{7}{15} n is an integer.

Solution

7. (i) (iv) Use Example 3 of §2 and Problem 2; (v) Use (iv) and Example 3 of §2; (vii) Use (vi), 5n5n5 \mid n^{5}-n, and Example 3 of §2; (viii) Similar to (vii); (ix) Use 5n5n,3n3n5\left|n^{5}-n, 3\right| n^{3}-n to deduce that there exists an integer AA such that
n5/5+n3/3+7n/15=n/5+n/3+7n/15+A=n+A.n^{5} / 5+n^{3} / 3+7 n / 15=n / 5+n / 3+7 n / 15+A=n+A .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.