INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A25

A QUESTION OF SIERPINSKI ON TRIANGULAR NUMBERS

Michael A. Bennett

Department of Mathematics, University of British Columbia, Vancouver BC, Canada

                           bennett@math.ubc.edu

Received: 8/30/05, Accepted: 11/6/05, Published: 11/22/05

Abstract

We answer a question of Sierpinski by showing that there do not exist four distinct triangular number in geometric progression.

In D23 of [2], it is stated that Sierpinski asked the question of whether or not there exist four (distinct) triangular numbers in geometric progression and, further, that Szymiczek [4] conjectured the answer to this to be a negative one. Recall that a triangular number is one of the form

T_{n} = \frac{n(n + 1)}{2}

for n ∈ ℤ. The problem of finding three such triangular numbers is readily reduced to finding solutions to a Pell equation (whereby, an old result of Gérardin [1] (see also [3]) implies that there are infinitely many such triples, the smallest of which is (T_{1}, T_{3}, T_{8})). It is easy to show that the answer to Sierpinski’s question is in the negative. This is, in fact, an immediate consequence of the following.

Lemma If a and b are positive integers with b > 1 then at least one of ab + 1 and ab^{3} + 1 is not a perfect square.

Proof. Suppose that we have

ab + 1 = x^{2} and ab^{3} + 1 = y^{2}

for positive integers x and y. From the theory of quadratic fields, since we may assume that ab is not a perfect square, it follows that

y + b\sqrt{ab} = (x + \sqrt{ab})^{k}

for some positive integer k (which, since b > 1, we may assume to be at least 2). If k ≥ 3, then we would have that

y + b\sqrt{ab} ≥ (x + \sqrt{ab})^{3}

and hence

b ≥ 3x^{2} + ab,

a contradiction, since a, x ≥ 1. It follows that k = 2 and hence that b = 2x. Since ab+1 = x^{2}, we have x = 1, contradicting the fact that a and b are positive integers. This completes the proof of our Lemma.

To connect this to Sierpinski’s question, suppose that the four triangular numbers in geometric progression are

T_{x} < T_{y} < T_{u} < T_{v}.

Taking a = 8T_{x} and b = T_{y}/T_{x}, it follows that one of 8T_{y} + 1 or 8T_{v} + 1 cannot be a perfect square, contradicting the identity 8T_{n} + 1 = (2n + 1)^{2}.

References

[1] A. Gérardin. Sphinx-Oedipe 9 (1914), 75, 145–146

[2] R. Guy. Unsolved Problems in Number Theory, 3rd edition. Springer Verlag, New York, 2004.

[3] K. Szimiczek. L’ équation uv = w^{2} en nombres triangulaires. (French) Publ. Inst. Math. (Beograd) (N.S.) 3 (17) (1963), 139–141.

[4] K. Szimiczek. The equation (x^{2} − 1)(y^{2} − 1) = (z^{2} − 1)^{2}, Eureka 35 (1972), 21–25.