Inspired by Naren Bhandari

Number Theory Level 3

$( {\color{#3D99F6} a } - 1 ) \times {\color{#3D99F6} a } \times ({\color{#3D99F6} a } + 1) = 6{\color{#D61F06}b}^2$

Clearly, $(a, b)=(2, 1)$ and $(3, 2)$ are solutions because $1 \times 2 \times 3 = 6 \times 1^2$ and $2 \times 3 \times 4 = 6 \times 2 ^ 2.$

If $a$ and $b$ are positive integers and $b\ge 3,$ is there a solution that satisfies the above equation?

Inspiration

No Yes

1 solution

David Vreken
May 5, 2018

One solution is $(a, b) = (49, 140)$ , since $(a - 1) \cdot a \cdot (a + 1) = (49 - 1) \cdot 49 \cdot (49 + 1) = 117600$ and $6b^2 = 6(140)^2 = 117600$ .

Let $n = a - 1$ . Then $n(n + 1)(n + 2) = 6b^2$ and $\frac{n(n + 1)(n + 2)}{6} = b^2$ . Since $\frac{n(n + 1)(n + 2)}{6}$ is the formula for the $n^{th}$ tetrahedral number, integer solutions for $a$ and $b$ occur when a tetrahedral number is also a square number. According to to this Wikipedia article , "A. J. Meyl proved in 1878 that only three tetrahedral numbers are also perfect squares, namely: $T_1 = 1^2 = 1$ , $T_2 = 2^2 = 4$ , and $T_{48} = 140^2 = 19600$ ." Since $a = n + 1$ , the only possible solutions occur when $a = 1 + 1 = 2$ , $a = 2 + 1 = 3$ , and $a = 48 + 1 = 49$ .

Do you know if there are (infinitely) more solutions?

There are several cases of how the "multiples of a perfect square" could line up, and 3 cases correspond to $(x^2, 2y^2, 3z^2), (2x^2, 3y^2, z^2), (3x^2, y^2, 2z^2)$ . In each of these cases, I cannot yet eliminate other possibilities.

I didn't like the $a = 2, 3$ solutions as they were small enough cases for someone to consider when answering this problem. I was wanting to come up with a less trivial variant of this version, but there doesn't really seem to be a nice one.

Calvin Lin Staff - 3 years, 1 month ago

I originally answered this by trial and error and got the a = 2 and a = 3 straightaway. I wrote a quick Python computer program and discovered that 2, 3, and 49 are the only solutions for a < 1,000,000 which suggests to me that there are finite solutions, but I do not know if this is true or how to prove it if it is.

I was just about to suggest that the question may be better if it limited a or b to solutions greater than (a, b) = (2, 1) and (a, b) = (3, 2), since (a, b) = (49, 140) is not as obvious, but I see you already did that.

David Vreken - 3 years, 1 month ago

Here is a link to the article, written in French.

It cites another paper (by the same journal) that deals with $a - 1$ being even leading only to solutions $a-1 = 2, 48$ .
It then proves the case of $a - 1$ being odd only has solutions when $a + 1$ is a multiple of 3, and even then the only solution requires $a - 1 = 1$ .

Calvin Lin Staff - 3 years, 1 month ago

According to this Wikipedia article , "A. J. Meyl proved in 1878 that only three tetrahedral numbers are also perfect squares, namely: $T_1 = 1^2 = 1$ , $T_2 = 2^2 = 4$ , and $T_{48} = 140^2 = 19600$ ." I have not been able to find the proof for this, but apparently it is quite long and complex.

David Vreken - 3 years, 1 month ago

Any approach on how to find the solution (49,140)?

Peter van der Linden - 3 years, 1 month ago

To be honest, I used a Python computer program.

David Vreken - 3 years, 1 month ago

Inspired by Naren Bhandari

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