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Problem statement:

Does there exist positive integers $a<b<c$ such that $$1 + 2 + \dots + (a-1) = (a+1) + \dots + (b-1) = (b+1) + \dots + c?$$ (Note that $a$ and $b$ are not in the sums.)

Motivation: this puzzle (some progress is made).

Notable progress: Gareth McCaughan proved using Pell's equations that if a solution exists, then $c>10^{600000}$. Techniques using elliptic curves were also developed.

Remark: This reminds of this easy-to-state Diophantine equation with large solutions (MO link).

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    $\begingroup$ is there a reason to omit $a$ and $b$ from the question (other than the original motivation)? $\endgroup$
    – Benjamin Dickman
    Commented Jun 21 at 16:18
  • $\begingroup$ @BenjaminDickman this question is equivalent to the motivation, so no, at the moment. $\endgroup$
    – Benjamin Wang
    Commented Jun 22 at 8:03

1 Answer 1

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This is just an equivalent formulation:

Find squares in the linear recurrence sequence (besides first two terms):

$$9, 25, 481, 14961, 500889, 16973353, \dotsc$$

which can be described equivalently by

  • recurrence $t_n = 41 t_{n-1} - 246 t_{n-2} + 246 t_{n-3} - 41 t_{n-4} + t_{n-5}$ for $n\geq 5$;
  • generating function $$\frac{9 - 344x + 1670x^2 - 824x^3 + 33x^4}{(1-x)(1-6x+x^2)(1-34x+x^2)}$$
  • explicit formula $$t_n = \frac38 \big((1+\sqrt2)^{4n} + (1-\sqrt2)^{4n}\big) + \frac{4-\sqrt2}2 (1+\sqrt2)^{2n} + \frac{4+\sqrt2}2 (1-\sqrt2)^{2n} + \frac{17}4.$$

Suppose that $t_n=z^2$ for some integer $z$, then values of $a,b,c$ are given by $$\begin{cases} a = \frac{(1+\sqrt2)^{2n} - (1-\sqrt2)^{2n}}{4\sqrt2}; \\ b = \frac{(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n} + 2}{4}; \\ c = \frac{z-1}2. \end{cases}$$

I believe it is a hard question to determine squares in linear recurrences like that. It is more or less solved only for recurrences of the second order (like Fibonacci numbers).

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    $\begingroup$ It's hard to find squares in the recurrence, but to lower bound them it's quite straightforward to test for quadratic residues against a number of small primes. Using the $24$ largest primes under $10^6$ for a first filter and then testing any which pass that against odd primes from 3 until rejection, with 3 hours of runtime I calculate that $n > 8000000000$, which gives a lower bound for $c$ on the order of $10^{12248800000}$. $\endgroup$
    – Peter Taylor
    Commented Jun 22 at 11:30

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