For each positive integer n, determine the least integer m such that lcm{1, 2

Problem 11761

(American Mathematical Monthly, Vol.121, March 2014) Proposed by B. Tomper (USA).

For each positive integer n, determine the least integer m such that lcm{1, 2, . . . , m} = lcm{n, n + 1, . . . , m}.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Let a1= 1, a2= 2, and, for n > 2, let an be twice the largest power of a prime less than n.

The first terms of this sequence are

1, 2, 4, 6, 8, 10, 10, 14, 16, 18, 18, 22, 22, 26, 26, 26, 32, . . . . We claim that the least integer m with the required property is just an.

Let n > 2, and let an = 2p^α, then p^α≥ 2, and, by Bertrand’s postulate, there is a prime q ∈ (p^α, 2p^α).

It follows that 2p^α≥ n, otherwise p^α< q < 2p^α< n, contradicting the fact that p^αis largest power of a prime less than n. Moreover, by definition of lcm,

lcm{1, 2, . . . , m} = Y

p prime

pbln(m)/ ln(p)c≥ lcm{n, n + 1, . . . , m}

and when equality holds for some m0 then equality is verified also for all m ≥ m0. If m < an= 2p^α, then p^α< n ≤ m < 2p^α implies that

p^α| lcm{1, 2, . . . , m}, but p^α6 | lcm{n, n + 1, . . . , m}

so lcm{1, 2, . . . , m} > lcm{n, n + 1, . . . , m}.

It remains to show that if m = an= 2p^α then

lcm{1, 2, . . . , 2p^α} = lcm{n, n + 1, . . . , 2p^α}.

By the pigeonhole principle, any number in {1, 2, . . . , p^α} has at least one multiple in the set {p^α+ 1, p^α+ 2, . . . , 2p^α} (which has p^α elements). Hence

lcm{1, 2, . . . , 2p^α} = lcm{p^α+ 1, p^α+ 2, . . . , 2p^α}.

In order to prove that

lcm{p^α+ 1, p^α+ 2, . . . , 2p^α} = lcm{n, n + 1, . . . , 2p^α}

it suffices to show that if q^βdivides lcm{p^α+1, p^α+2, . . . , n−1} for some prime q, then q^β≤ 2p^α−n+1 which implies, by the pigeonhole principle, that there is a multiple of q^βin the set {n, n+1, . . . , 2p^α}.

It is easy to verify it holds for n ≤ 26. If n ≥ 26 then p^α≥ 25 and by a stronger version of Bertrand’s postulate proved by J. Nagura in On the interval containing at least one prime number (1952), there is a prime between p^αand 6p^α/5. As above, this implies that 6p^α/5 ≥ n. Since p^αis largest power of a prime less than n, it follows that there is an integer b ≥ 2 such that bq^β ∈ {p^α+1, p^α+2, . . . , n−1}.

Thus n > 2q^β and

2p^α− n + 1 > 2p^α−6p^α 5 = 4p^α

5 ≥ 4 5

5n 6 = 2n

3 >n 2 > q^β,

and the proof is complete.