Can the sum of the first $p$ factorials ever be a perfect power for $ p>3 $?
$begingroup$
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!
linear-algebra factorial
edited Jan 19 at 14:29
Peter
47.6k1039131
$endgroup$
add a comment |
$begingroup$
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!
linear-algebra factorial
edited Jan 19 at 14:29
Peter
47.6k1039131
$endgroup$
add a comment |
$begingroup$
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!
linear-algebra factorial
edited Jan 19 at 14:29
Peter
47.6k1039131
$endgroup$
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!
linear-algebra factorial
linear-algebra factorial
edited Jan 19 at 14:29
Peter
47.6k1039131
edited Jan 19 at 14:29
Peter
47.6k1039131
edited Jan 19 at 14:29
Peter
47.6k1039131
edited Jan 19 at 14:29
Peter
47.6k1039131
edited Jan 19 at 14:29
Peter
47.6k1039131
47.6k1039131
asked Jan 19 at 14:13
user636268
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $S(8)=3^2times 11times 467$ 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 $p$).
answered Jan 19 at 14:40
lulululu
41.6k24979
$endgroup$
1
$begingroup$
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
1
$begingroup$
@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
2
$begingroup$
@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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $S(8)=3^2times 11times 467$ 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 $p$).
answered Jan 19 at 14:40
lulululu
41.6k24979
$endgroup$
1
$begingroup$
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
1
$begingroup$
@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
2
$begingroup$
@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
add a comment |
$begingroup$
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 $S(8)=3^2times 11times 467$ 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 $p$).
answered Jan 19 at 14:40
lulululu
41.6k24979
$endgroup$
1
$begingroup$
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
1
$begingroup$
@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
2
$begingroup$
@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
add a comment |
$begingroup$
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 $S(8)=3^2times 11times 467$ 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 $p$).
answered Jan 19 at 14:40
lulululu
41.6k24979
$endgroup$
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 $S(8)=3^2times 11times 467$ 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 $p$).
answered Jan 19 at 14:40
lulululu
41.6k24979
answered Jan 19 at 14:40
lulululu
41.6k24979
answered Jan 19 at 14:40
lulululu
41.6k24979
answered Jan 19 at 14:40
lulululu
41.6k24979
41.6k24979
1
$begingroup$
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
1
$begingroup$
@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
2
$begingroup$
@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
add a comment |
1
$begingroup$
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
1
$begingroup$
@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
2
$begingroup$
@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
1
1
$begingroup$
You answered despite of the demanding style ?
$endgroup$
– Peter
Jan 19 at 14:43
$begingroup$
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
$begingroup$
@Peter I didn't look at the comments, you are right about the tone.
$endgroup$
– lulu
Jan 19 at 14:44
1
1
$begingroup$
@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
$begingroup$
@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
2
2
$begingroup$
@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
$begingroup$
@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
$begingroup$
@lulu I think you are right, and the elegant solution is surely useful also for other users.
$endgroup$
– Peter
Jan 19 at 15:23
add a comment |
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