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Suppose that $a,n,t$ are positive integers greater than $1$ and $q$ is a prime number. I write $a_{n} = \binom {a^{n}} {a}$. In general I was curious about $a_{n}-k =q^{t}$ in particular the case $a=2$. Now quickly I can show that \begin{align}2_{n}=\frac{2^{n}(2^{n}-1)(2^{n}-2)!}{2(2^{n}-2)!} = 2^{n-1}(2^{n}-1) \end{align}

So I am really looking at $2^{n-1}(2^{n}-1)-k =q^{t}$. Is this an exponential Diophantine equation? If so what are the possible solutions $(n,k,q,t)$ that solve that equation?

Here is what I have so far: $(2,2,2,2)$, $(3,3,5,2)$, $(4,39,3,4)$, and $(5,135,19,2)$ are solutions. In particular $q=2,3,5$ and $19$.

I believe that $k$ has to be an odd number whenever $n>2$. This can be seen with a parity argument. Certainly $2^{n-1}(2^{n}-1)$ is even and $q^{t}$ is odd. Then $EVEN -k = ODD$. So $K$ is odd.

The examples I have are by hand. I have not written any code to search for a bigger pattern. I do think this equation has a solution for every integer $n$.

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  • $\begingroup$ Of course that are there an infinite number of solutions . Take $n,q,t$ arbitrary with the condition that $2_n>q^t$ and then choose : $$k=2_n-q^t$$ $\endgroup$
    – user252450
    Mar 23, 2016 at 19:10
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    $\begingroup$ For each $n$ you take all the powers of primes less than $2_n$ and you have obviuos solutions for each of these cases. $\endgroup$
    – Piquito
    Mar 23, 2016 at 19:13
  • $\begingroup$ @Piquito Nailed it and obvious, thanks! $\endgroup$ Mar 23, 2016 at 19:21

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See @Piquito's comment. Pick any power of a prime less than $2_{n}$ and you have a solution. So for example take $2_{6}=2016$. Now $7^{2}<2016$ and $2016-49=1967$. So we have that $2_{6}-1967 = 7^{2}$

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