Number theoryDifficulty 6.0National olympiadProve it
7. Prove that for any integer n, we have (i) 6∣n(n+1)(n+2); (ii) 8∣n(n+1)(n+2)(n+3); (iii) 24∣n(n+1)(n+2)(n+3); (iv) if 2∤n, then 8∣n2−1 and 24∣n(n2−1); (v) if 2∤n and 3∤n, then 24∣n2+23; (vi) 6∣n3−n; (vii) 30∣n5−n; (viii) 42∣n7−n; (ix) 51n5+31n3+157n 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), 5∣n5−n, and Example 3 of §2; (viii) Similar to (vii); (ix) Use 5n5−n,3n3−n to deduce that there exists an integer A such that n5/5+n3/3+7n/15=n/5+n/3+7n/15+A=n+A.
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