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Has $$\sum_{j=1}^p j!=q^r$$ , where q,p,r are positive integers, and r > 1 , a solution ?

I solved partially, if r is even, then RHS is a perfect square, and there is no doubt in that. Therefore, the left side must be a perfect square as well. But for p>3, the last digit of LHS is 3 which is not the last digit of any square. Hence, p<4, and hence manually checking, only solutions are p =1,3. But i cannot generalize when r is odd. Any solution to the next part would be helpful!

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Let $S(p)=\sum_{j=1}^pj!$ denote your sum.

Claim: For $p≥8$ $v_3(S(p))=2$

(Here, as usual, for a prime $q$ and natural number $n$, $v_q(n)$ denotes the order to which $q$ divides $n$. Thus, $v_3(18)=2$ for example.).

Pf: One computes that $S(8)=3^2\times 11\times 467$ and from there after you have at least three factors of $3$ in each summand.

It follows immediately that, at least for $p≥8$ your sum could not be any power of degree greater than $2$. As you have already addressed the case of squares, we are done (after a simple search for small $p$).

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    $\begingroup$ You answered despite of the demanding style ? $\endgroup$ – Peter Jan 19 at 14:43
  • $\begingroup$ @Peter I didn't look at the comments, you are right about the tone. $\endgroup$ – lulu Jan 19 at 14:44
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    $\begingroup$ @lulu I think users should not be encouraged to get answers this way. So, my suggestion is to delete the answer , although it is a very nice answer. $\endgroup$ – Peter Jan 19 at 14:50
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    $\begingroup$ @Peter I'm on the fence. The question itself is very well posed, I thought. The OP addresses a major subcase of the problem. The tone of the comments is bad, and had I read them I would have stopped thinking about the problem. As it is, I think the question is mathematically interesting so I'm going to leave the solution up, though I do think you have a point. $\endgroup$ – lulu Jan 19 at 15:22
  • $\begingroup$ @lulu I think you are right, and the elegant solution is surely useful also for other users. $\endgroup$ – Peter Jan 19 at 15:23

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