# nt.number theory – Products equal to consecutive integers

Summary: What are the non-trivial solutions to the question? Find two consecutive integer sequences whose products are equal. There are four known solutions, all of which consist of small integers, and there is no clear pattern connecting them. Given the small number of known solutions and the lack of a clear pattern between them, I suspect that this question contains more number theory than it first appears.

The "puzzle corner" of MIT News For March / April 2020, Sorab Vatcha poses a "speed" problem: "Find seven consecutive integers whose product equals the product of four consecutive integers." The obvious solution to "speed" is: 0, 1, 2, 3, 4, 5, and 6; and 0, 1, 2 and 3.

This leads to the general question of whether there are "nontrivial" solutions to find two different sequences of integers whose products are the same. Obvious trivial solutions involve (1) sequences containing 0, (2) replacing all integers in a sequence with their negatives, (3) adding or removing values ​​of 1 or -1, and (4) a sequence having length 1 (and therefore it contains an integer that is the product of the integers in the other sequence.) These criteria reduce the problem to finding two different sequences of integers $$ge 2$$ of length $$ge 2$$ whose products are the same.

There is a less obvious criterion of non-triviality: (5) The two sequences must not overlap. If the sequences overlap, then the overlapping part can be removed from both, producing a solution with shorter sequences. This interacts with prohibition (4) against length 1 sequences: removing the overlapping part can reduce a sequence to length 1, and the shorter solution can also be trivial. And, in fact, there is a large family of solutions built in this way: if the product of $$a cdots b$$ it is $$P$$, then $$a cdots (P-1) = (b + 1) cdots P$$.

There are non-trivial solutions. The one with the smallest product is $$5 cdot 6 cdot 7 = 14 cdot 15 = 210$$. Is there an enumeration of all solutions?

The known non-trivial solutions are:
$$5 cdots 7 = 14 cdot 15 = 210$$,
$$2 cdots 6 = 8 cdots 10 = 720$$,
$$19 cdots 22 = 55 cdots 57 = 175,560$$Y
$$8 cdots 14 = 63 cdots 66 = 17297280$$.

I have conducted several computer searches that have revealed no other solution: (a) all sequences with products less than $$10 30$$, (b) sequences with numbers less than 10,000,000 and length less than 10, (c) sequences with numbers less than 1,000,000 and length less than 100, (d) sequences with numbers less than 100,000 and length less than 1,000, and (e) sequences with numbers less than 10,000 and length less than 10,000.

Also see https://math.stackexchange.com/questions/991728/equal-products-of-consecutive-integers/ and https://math.stackexchange.com/questions/3346618/non-trivial-solutions-to-equal -secutive-integer-products.