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The short version of my question is:

1)For which positive integers $k, n$ is there a solution to the equation $$k(6k+1)=1+q+q^2+\cdots+q^n$$ with $q$ a prime power?

2) For which positive integers $k, n$ is there a solution to the equation $$(3k+1)(2k+1)=1+q+q^2+\cdots+q^n$$ with $q$ a prime power?

Now for some motivation. In this question I ask for an algebra-geometric construction of certain Steiner systems (there's background on what Steiner systems are and on the details and motivation for the construction in that question). In particular, if the construction can be carried out for a $(p, q, r)$ Steiner system, the blocks of the Steiner system would be given by the $p-1$-plane sections of some variety $X$ in affine or projective space over a finite field $\mathbb{F}_q$. If $X$ is in affine $n+1$-space and $p=2$, the number of blocks would be given by $1+q+\cdots+q^n$. Now the number of blocks in a $(2, 3, 6k+1)$ system is $k(6k+1)$ and the number of blocks in a $(2, 3, 6k+3)$ system is $(3k+1)(2k+1)$ (a $(2, 3, n)$ Steiner system is realizable iff $n=1, 3$ mod $6$, $n > 3$). So the question comes from setting these two quantities equal.

In other words, these equations must be solvable for all $k$ if the algebra-geometric construction of Steiner systems is to go through for all $(2, 3, n)$ systems (the Steiner triple systems) in affine space.

The most general form of this question (which covers both the affine and the projective versions) is:

For integers $n, 1 < p < q < r$, when is there a prime power $q$ such that $$\frac{r!(q-p)!}{q!(r-p)!}=\left[n \atop p-1\right]_q$$ or $$\frac{r!(q-p)!}{q!(r-p)!}=\left[n \atop p\right]_q?$$

Of course, I only expect answers to concentrate on the numbered questions (1) and (2) at the top of the page.

EDIT: Note that e.g. $k=4$ has no solutions for the first equation.

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I can only comment that you have weird tags but not the right one, diophantine equations. –  Wadim Zudilin Jun 26 '10 at 3:55
    
Fixed; tag added. –  Daniel Litt Jun 26 '10 at 6:26
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1 Answer

up vote 5 down vote accepted

Let $t = 1+q+q^2+\dots+q^n $ then each of the equations (1) and (2) implies that $24t+1$ is a square (namely, $24t+1=(12k+1)^2$ and $24t+1=(12k+5)^2$, respectively). For $n=2$ that leads to a Pellian equation (with possibly infinitely many solutions), for $n=3,4$ to an elliptic curve (with finitely many solutions, if any), and for $n>4$ to a hyper-elliptic curve (with no solutions for most $n$).

Cases $n=3,4$ are easy to solve.

For $n=3$, integer solutions are $q=-1, 0, 2, 3, 13, 25, 32, 104, 177$ out of which only $2,3,13,25,32$ are powers of primes. For $n=4$, integral solutions are $q=-1,0,1,25,132$ out of which only $25$ is a power of prime. These numerical values are computed in SAGE and MAGMA.

Also, for a fixed value of $k$, it is possible to verify solubility of the given equations by iterating all possible $q$ dividing the l.h.s. minus 1. In particular, equation (1) has solutions only for the following $k$ below $10^6$: 1, 2, 3, 15, 52, 75, 1302, 32552, 813802. Similarly, equation (2) has solutions only for the following $k$ below $10^6$: 1, 10, 260, 6510, 162760.

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Thanks! I have to refresh my number theory. I was following the strategy in your last paragraph, but didn't have mathematical software on hand so stopped at $k=4$; this seems to suggest that any algebra-geometric construction of general Steiner systems is going to be pretty complicated if it exists. –  Daniel Litt Jun 27 '10 at 7:03
    
Max, this is a nice argument! –  Wadim Zudilin Jun 29 '10 at 0:49
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