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Question

# Find all the natural numbers n such that n2 does not divide (n−2)! .

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Solution

## Suppose n=pqr, where p<q are primes and r>1.Then, p≤2,q≥3 and r≥2, not necessarily a prime. Thus, we haven−2≥n−p=pqr−p≥5p>pn−2≥n−q=q(pr−1)≥3q>qn−2≥n−pr=pr(q−1)≥2pr>prn−2≥n−qr=qr(p−1)≥qrObserve that p,q,pr,qr are all distinct. Hence, there product divides (n−2) !. Thus, n2=p2q2r2 devides (n−2)! In this case, n=pq where p,q are distinct primes n=pk for some p.Case 1. Suppose n=pq for some primes p,q where 2<p<q. Then, p≥3 and q≥5 .In this case,n−2>n−p=p(q−1)≥4pn−2>n−q=q(p−1)≥2qThus, p,q,2p,2q are all distinct numbers in the set {1, 2, 3, ...., n - 2}. We see that n2=p2q2 divides (n−2)!. We conclude that n=2q for some prime numbers q≥3. Note that, n−2=2q−2<2q in this case so that n2 does not divide (n−2)!Case 2. Suppose n=kp for some prime p. We observe that p,2p,3p,...,(pk−1−1)p all lie in the set {1, 2, 3, ...,n−2 }. If pk−1−1≥2k, then there are at least 2k multiples of p in the set {1, 2, 3, ..., n−2 }. Hence, n2=p2k divides (n−2)!. Thus, pk−1−1<2k. If k≥5, then pk−1−1≥2k−1−1≥2k, which maybe proved by an easy induction. Hence, k≤4. If k=1, we get n=p, a prime. If k=2, then p−1<4 so that p=2 or 3; we get n=22=4 or n=32=9. For k=3, we have p2−1<6 giving p=2; n=23=8 in this case. Finally , k=4 gives p3−1<8. Again, p=2 and n=24=16. However, n2=28 divides 14! and hence it is not a solution.Thus, n=2,2p for some prime p or n=8,9. It is easy to verify that these satisfy the conditions of the problem.

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