Mixing
prove that [p!/(p-4)!] + 1 is a perfect square for all natural p.
$begingroup$
one can observe that $[p!/(p-4)!] + 1$ is basically the product of four consecutive integers plus one.Since this is
$$
begin{eqnarray} p(p+1)(p+2)(p+3)+1 & = &(p^2+3p)(p^2+3p+2)+1 \
& = & [(p^2+3p+1)−1][(p^2+3p+1)+1]+1 \
& = & (p^2+3p+1)^2 end{eqnarray} $$
which is a perfect square for all natural $p$ but my teacher says that there is a more elegant method of doing it, help in any form would be appreciated.
factorial factoring square-numbers
asked Jan 28 at 14:12
Mary TomMary Tom
326
$endgroup$
|
show 5 more comments
$begingroup$
one can observe that $[p!/(p-4)!] + 1$ is basically the product of four consecutive integers plus one.Since this is
$$
begin{eqnarray} p(p+1)(p+2)(p+3)+1 & = &(p^2+3p)(p^2+3p+2)+1 \
& = & [(p^2+3p+1)−1][(p^2+3p+1)+1]+1 \
& = & (p^2+3p+1)^2 end{eqnarray} $$
which is a perfect square for all natural $p$ but my teacher says that there is a more elegant method of doing it, help in any form would be appreciated.
factorial factoring square-numbers
asked Jan 28 at 14:12
Mary TomMary Tom
326
$endgroup$
$begingroup$
Did you say n when you meant to say p in your title?
$endgroup$
– Mark
Jan 28 at 14:19
$begingroup$
Welcome to MSE! Did you mean for all natural $p$?
$endgroup$
– J. W. Tanner
Jan 28 at 14:19
1
$begingroup$
Are you sure that it is $p(p+1)(p+2)(p+3)+1$ and not $p(p-1)(p-2)(p-3)+1$?
$endgroup$
– user 170039
Jan 28 at 14:20
1
$begingroup$
You are using $p$ and $n$ in confusing ways, Your target seems to be the product of four consecutive integers plus one. But these integers are named inconsistently.
$endgroup$
– Mark Bennet
Jan 28 at 14:24
3
$begingroup$
Your computation appears to be working with $frac {(p+4)!}{p!}+1$ instead of what you wrote. Doesn't change much, but it makes your post unhelpfully confusing.
$endgroup$
– lulu
Jan 28 at 14:24
|
show 5 more comments
$begingroup$
one can observe that $[p!/(p-4)!] + 1$ is basically the product of four consecutive integers plus one.Since this is
$$
begin{eqnarray} p(p+1)(p+2)(p+3)+1 & = &(p^2+3p)(p^2+3p+2)+1 \
& = & [(p^2+3p+1)−1][(p^2+3p+1)+1]+1 \
& = & (p^2+3p+1)^2 end{eqnarray} $$
which is a perfect square for all natural $p$ but my teacher says that there is a more elegant method of doing it, help in any form would be appreciated.
factorial factoring square-numbers
asked Jan 28 at 14:12
Mary TomMary Tom
326
$endgroup$
one can observe that $[p!/(p-4)!] + 1$ is basically the product of four consecutive integers plus one.Since this is
$$
begin{eqnarray} p(p+1)(p+2)(p+3)+1 & = &(p^2+3p)(p^2+3p+2)+1 \
& = & [(p^2+3p+1)−1][(p^2+3p+1)+1]+1 \
& = & (p^2+3p+1)^2 end{eqnarray} $$
which is a perfect square for all natural $p$ but my teacher says that there is a more elegant method of doing it, help in any form would be appreciated.
factorial factoring square-numbers
factorial factoring square-numbers
asked Jan 28 at 14:12
Mary TomMary Tom
326
asked Jan 28 at 14:12
Mary TomMary Tom
326
edited Jan 28 at 14:25
Mary Tom
asked Jan 28 at 14:12
Mary TomMary Tom
326
asked Jan 28 at 14:12
Mary TomMary Tom
326
asked Jan 28 at 14:12
Mary TomMary Tom
326
326
$begingroup$
Did you say n when you meant to say p in your title?
$endgroup$
– Mark
Jan 28 at 14:19
$begingroup$
Welcome to MSE! Did you mean for all natural $p$?
$endgroup$
– J. W. Tanner
Jan 28 at 14:19
1
$begingroup$
Are you sure that it is $p(p+1)(p+2)(p+3)+1$ and not $p(p-1)(p-2)(p-3)+1$?
$endgroup$
– user 170039
Jan 28 at 14:20
1
$begingroup$
You are using $p$ and $n$ in confusing ways, Your target seems to be the product of four consecutive integers plus one. But these integers are named inconsistently.
$endgroup$
– Mark Bennet
Jan 28 at 14:24
3
$begingroup$
Your computation appears to be working with $frac {(p+4)!}{p!}+1$ instead of what you wrote. Doesn't change much, but it makes your post unhelpfully confusing.
$endgroup$
– lulu
Jan 28 at 14:24
|
show 5 more comments
$begingroup$
Did you say n when you meant to say p in your title?
$endgroup$
– Mark
Jan 28 at 14:19
$begingroup$
Welcome to MSE! Did you mean for all natural $p$?
$endgroup$
– J. W. Tanner
Jan 28 at 14:19
1
$begingroup$
Are you sure that it is $p(p+1)(p+2)(p+3)+1$ and not $p(p-1)(p-2)(p-3)+1$?
$endgroup$
– user 170039
Jan 28 at 14:20
1
$begingroup$
You are using $p$ and $n$ in confusing ways, Your target seems to be the product of four consecutive integers plus one. But these integers are named inconsistently.
$endgroup$
– Mark Bennet
Jan 28 at 14:24
3
$begingroup$
Your computation appears to be working with $frac {(p+4)!}{p!}+1$ instead of what you wrote. Doesn't change much, but it makes your post unhelpfully confusing.
$endgroup$
– lulu
Jan 28 at 14:24
$begingroup$
Did you say n when you meant to say p in your title?
$endgroup$
– Mark
Jan 28 at 14:19
$begingroup$
Did you say n when you meant to say p in your title?
$endgroup$
– Mark
Jan 28 at 14:19
$begingroup$
Welcome to MSE! Did you mean for all natural $p$?
$endgroup$
– J. W. Tanner
Jan 28 at 14:19
$begingroup$
Welcome to MSE! Did you mean for all natural $p$?
$endgroup$
– J. W. Tanner
Jan 28 at 14:19
1
1
$begingroup$
Are you sure that it is $p(p+1)(p+2)(p+3)+1$ and not $p(p-1)(p-2)(p-3)+1$?
$endgroup$
– user 170039
Jan 28 at 14:20
$begingroup$
Are you sure that it is $p(p+1)(p+2)(p+3)+1$ and not $p(p-1)(p-2)(p-3)+1$?
$endgroup$
– user 170039
Jan 28 at 14:20
1
1
$begingroup$
You are using $p$ and $n$ in confusing ways, Your target seems to be the product of four consecutive integers plus one. But these integers are named inconsistently.
$endgroup$
– Mark Bennet
Jan 28 at 14:24
$begingroup$
You are using $p$ and $n$ in confusing ways, Your target seems to be the product of four consecutive integers plus one. But these integers are named inconsistently.
$endgroup$
– Mark Bennet
Jan 28 at 14:24
3
3
$begingroup$
Your computation appears to be working with $frac {(p+4)!}{p!}+1$ instead of what you wrote. Doesn't change much, but it makes your post unhelpfully confusing.
$endgroup$
– lulu
Jan 28 at 14:24
$begingroup$
Your computation appears to be working with $frac {(p+4)!}{p!}+1$ instead of what you wrote. Doesn't change much, but it makes your post unhelpfully confusing.
$endgroup$
– lulu
Jan 28 at 14:24
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Only thing I can see:
Note that $p(p-3)=p^2-3p$ and $(p-1)(p-2)=p^2-3p+2$. It follows that we can write your expression as
$$(p^2-3p)^2+2(p^2-3p)+1=m^2+2m+1=(m+1)^2$$
Where we have introduced $m=p^2-3p$ ,
answered Jan 28 at 14:23
lulululu
43.2k25080
$endgroup$
$begingroup$
some of my classmates have done it like that.
$endgroup$
– Mary Tom
Jan 28 at 14:24
$begingroup$
This is pretty much the same as has been done in the question with a change of sign. Introducing $m$ makes it a little clearer.
$endgroup$
– Mark Bennet
Jan 28 at 14:26
$begingroup$
is there any other way?
$endgroup$
– Mary Tom
Jan 28 at 14:27
$begingroup$
@MarkBennet Maybe so. The point I was trying to make (maybe unsuccessfully, and maybe it wasn't all that good a point to begin with) was that it all hinges on the fact that $p^2-3p$ appears twice in the product. If it's unhelpful, I'll gladly delete.
$endgroup$
– lulu
Jan 28 at 14:28
$begingroup$
@lulu no, helpful, definitely.
$endgroup$
– Mark Bennet
Jan 28 at 14:32
add a comment |
$begingroup$
Slightly different approach... where $p=q+3$
$$S_q=frac{(p)!}{(p-4)!}+1=frac{(q+3)!}{(q-1)!} + 1=q(q+1)(q+2)(q+3)+1$$
Let $q=a-frac{3}{2}$
Then
$$S_q=left(a-frac{3}{2}right)left(a-frac{1}{2}right)left(a+frac{1}{2}right)left(a+frac{3}{2}right)+1$$
$$S_q=left(a^2-frac{3^2}{2^2}right)left(a^2-frac{1}{2^2}right)+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{3^2}{4^2}+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{5^2}{4^2}$$
$$S_q=left(a^2-frac{5}{4} right)^2$$
$$S_q=left((q+3/2)^2-frac{5}{4} right)^2$$
$$S_p=left((p-3/2)^2-frac{5}{4} right)^2$$
answered Jan 28 at 15:07
James ArathoonJames Arathoon
1,608423
$endgroup$
$begingroup$
I think this is what my teacher expected.
$endgroup$
– Mary Tom
Jan 28 at 15:38
$begingroup$
@MaryTom It then takes just a little bit of thought to know that if an integer is the square of a rational number, it is the square of an integer. Or to expand the expression in the bracket and see that it is an integer. It is not quite immediately obvious that the expression in the final bracket is an integer.
$endgroup$
– Mark Bennet
Jan 28 at 16:51
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Only thing I can see:
Note that $p(p-3)=p^2-3p$ and $(p-1)(p-2)=p^2-3p+2$. It follows that we can write your expression as
$$(p^2-3p)^2+2(p^2-3p)+1=m^2+2m+1=(m+1)^2$$
Where we have introduced $m=p^2-3p$ ,
answered Jan 28 at 14:23
lulululu
43.2k25080
$endgroup$
$begingroup$
some of my classmates have done it like that.
$endgroup$
– Mary Tom
Jan 28 at 14:24
$begingroup$
This is pretty much the same as has been done in the question with a change of sign. Introducing $m$ makes it a little clearer.
$endgroup$
– Mark Bennet
Jan 28 at 14:26
$begingroup$
is there any other way?
$endgroup$
– Mary Tom
Jan 28 at 14:27
$begingroup$
@MarkBennet Maybe so. The point I was trying to make (maybe unsuccessfully, and maybe it wasn't all that good a point to begin with) was that it all hinges on the fact that $p^2-3p$ appears twice in the product. If it's unhelpful, I'll gladly delete.
$endgroup$
– lulu
Jan 28 at 14:28
$begingroup$
@lulu no, helpful, definitely.
$endgroup$
– Mark Bennet
Jan 28 at 14:32
add a comment |
$begingroup$
Only thing I can see:
Note that $p(p-3)=p^2-3p$ and $(p-1)(p-2)=p^2-3p+2$. It follows that we can write your expression as
$$(p^2-3p)^2+2(p^2-3p)+1=m^2+2m+1=(m+1)^2$$
Where we have introduced $m=p^2-3p$ ,
answered Jan 28 at 14:23
lulululu
43.2k25080
$endgroup$
$begingroup$
some of my classmates have done it like that.
$endgroup$
– Mary Tom
Jan 28 at 14:24
$begingroup$
This is pretty much the same as has been done in the question with a change of sign. Introducing $m$ makes it a little clearer.
$endgroup$
– Mark Bennet
Jan 28 at 14:26
$begingroup$
is there any other way?
$endgroup$
– Mary Tom
Jan 28 at 14:27
$begingroup$
@MarkBennet Maybe so. The point I was trying to make (maybe unsuccessfully, and maybe it wasn't all that good a point to begin with) was that it all hinges on the fact that $p^2-3p$ appears twice in the product. If it's unhelpful, I'll gladly delete.
$endgroup$
– lulu
Jan 28 at 14:28
$begingroup$
@lulu no, helpful, definitely.
$endgroup$
– Mark Bennet
Jan 28 at 14:32
add a comment |
$begingroup$
Only thing I can see:
Note that $p(p-3)=p^2-3p$ and $(p-1)(p-2)=p^2-3p+2$. It follows that we can write your expression as
$$(p^2-3p)^2+2(p^2-3p)+1=m^2+2m+1=(m+1)^2$$
Where we have introduced $m=p^2-3p$ ,
answered Jan 28 at 14:23
lulululu
43.2k25080
$endgroup$
Only thing I can see:
Note that $p(p-3)=p^2-3p$ and $(p-1)(p-2)=p^2-3p+2$. It follows that we can write your expression as
$$(p^2-3p)^2+2(p^2-3p)+1=m^2+2m+1=(m+1)^2$$
Where we have introduced $m=p^2-3p$ ,
answered Jan 28 at 14:23
lulululu
43.2k25080
answered Jan 28 at 14:23
lulululu
43.2k25080
answered Jan 28 at 14:23
lulululu
43.2k25080
answered Jan 28 at 14:23
lulululu
43.2k25080
43.2k25080
$begingroup$
some of my classmates have done it like that.
$endgroup$
– Mary Tom
Jan 28 at 14:24
$begingroup$
This is pretty much the same as has been done in the question with a change of sign. Introducing $m$ makes it a little clearer.
$endgroup$
– Mark Bennet
Jan 28 at 14:26
$begingroup$
is there any other way?
$endgroup$
– Mary Tom
Jan 28 at 14:27
$begingroup$
@MarkBennet Maybe so. The point I was trying to make (maybe unsuccessfully, and maybe it wasn't all that good a point to begin with) was that it all hinges on the fact that $p^2-3p$ appears twice in the product. If it's unhelpful, I'll gladly delete.
$endgroup$
– lulu
Jan 28 at 14:28
$begingroup$
@lulu no, helpful, definitely.
$endgroup$
– Mark Bennet
Jan 28 at 14:32
add a comment |
$begingroup$
some of my classmates have done it like that.
$endgroup$
– Mary Tom
Jan 28 at 14:24
$begingroup$
This is pretty much the same as has been done in the question with a change of sign. Introducing $m$ makes it a little clearer.
$endgroup$
– Mark Bennet
Jan 28 at 14:26
$begingroup$
is there any other way?
$endgroup$
– Mary Tom
Jan 28 at 14:27
$begingroup$
@MarkBennet Maybe so. The point I was trying to make (maybe unsuccessfully, and maybe it wasn't all that good a point to begin with) was that it all hinges on the fact that $p^2-3p$ appears twice in the product. If it's unhelpful, I'll gladly delete.
$endgroup$
– lulu
Jan 28 at 14:28
$begingroup$
@lulu no, helpful, definitely.
$endgroup$
– Mark Bennet
Jan 28 at 14:32
$begingroup$
some of my classmates have done it like that.
$endgroup$
– Mary Tom
Jan 28 at 14:24
$begingroup$
some of my classmates have done it like that.
$endgroup$
– Mary Tom
Jan 28 at 14:24
$begingroup$
This is pretty much the same as has been done in the question with a change of sign. Introducing $m$ makes it a little clearer.
$endgroup$
– Mark Bennet
Jan 28 at 14:26
$begingroup$
This is pretty much the same as has been done in the question with a change of sign. Introducing $m$ makes it a little clearer.
$endgroup$
– Mark Bennet
Jan 28 at 14:26
$begingroup$
is there any other way?
$endgroup$
– Mary Tom
Jan 28 at 14:27
$begingroup$
is there any other way?
$endgroup$
– Mary Tom
Jan 28 at 14:27
$begingroup$
@MarkBennet Maybe so. The point I was trying to make (maybe unsuccessfully, and maybe it wasn't all that good a point to begin with) was that it all hinges on the fact that $p^2-3p$ appears twice in the product. If it's unhelpful, I'll gladly delete.
$endgroup$
– lulu
Jan 28 at 14:28
$begingroup$
@MarkBennet Maybe so. The point I was trying to make (maybe unsuccessfully, and maybe it wasn't all that good a point to begin with) was that it all hinges on the fact that $p^2-3p$ appears twice in the product. If it's unhelpful, I'll gladly delete.
$endgroup$
– lulu
Jan 28 at 14:28
$begingroup$
@lulu no, helpful, definitely.
$endgroup$
– Mark Bennet
Jan 28 at 14:32
$begingroup$
@lulu no, helpful, definitely.
$endgroup$
– Mark Bennet
Jan 28 at 14:32
add a comment |
$begingroup$
Slightly different approach... where $p=q+3$
$$S_q=frac{(p)!}{(p-4)!}+1=frac{(q+3)!}{(q-1)!} + 1=q(q+1)(q+2)(q+3)+1$$
Let $q=a-frac{3}{2}$
Then
$$S_q=left(a-frac{3}{2}right)left(a-frac{1}{2}right)left(a+frac{1}{2}right)left(a+frac{3}{2}right)+1$$
$$S_q=left(a^2-frac{3^2}{2^2}right)left(a^2-frac{1}{2^2}right)+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{3^2}{4^2}+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{5^2}{4^2}$$
$$S_q=left(a^2-frac{5}{4} right)^2$$
$$S_q=left((q+3/2)^2-frac{5}{4} right)^2$$
$$S_p=left((p-3/2)^2-frac{5}{4} right)^2$$
answered Jan 28 at 15:07
James ArathoonJames Arathoon
1,608423
$endgroup$
$begingroup$
I think this is what my teacher expected.
$endgroup$
– Mary Tom
Jan 28 at 15:38
$begingroup$
@MaryTom It then takes just a little bit of thought to know that if an integer is the square of a rational number, it is the square of an integer. Or to expand the expression in the bracket and see that it is an integer. It is not quite immediately obvious that the expression in the final bracket is an integer.
$endgroup$
– Mark Bennet
Jan 28 at 16:51
add a comment |
$begingroup$
Slightly different approach... where $p=q+3$
$$S_q=frac{(p)!}{(p-4)!}+1=frac{(q+3)!}{(q-1)!} + 1=q(q+1)(q+2)(q+3)+1$$
Let $q=a-frac{3}{2}$
Then
$$S_q=left(a-frac{3}{2}right)left(a-frac{1}{2}right)left(a+frac{1}{2}right)left(a+frac{3}{2}right)+1$$
$$S_q=left(a^2-frac{3^2}{2^2}right)left(a^2-frac{1}{2^2}right)+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{3^2}{4^2}+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{5^2}{4^2}$$
$$S_q=left(a^2-frac{5}{4} right)^2$$
$$S_q=left((q+3/2)^2-frac{5}{4} right)^2$$
$$S_p=left((p-3/2)^2-frac{5}{4} right)^2$$
answered Jan 28 at 15:07
James ArathoonJames Arathoon
1,608423
$endgroup$
$begingroup$
I think this is what my teacher expected.
$endgroup$
– Mary Tom
Jan 28 at 15:38
$begingroup$
@MaryTom It then takes just a little bit of thought to know that if an integer is the square of a rational number, it is the square of an integer. Or to expand the expression in the bracket and see that it is an integer. It is not quite immediately obvious that the expression in the final bracket is an integer.
$endgroup$
– Mark Bennet
Jan 28 at 16:51
add a comment |
$begingroup$
Slightly different approach... where $p=q+3$
$$S_q=frac{(p)!}{(p-4)!}+1=frac{(q+3)!}{(q-1)!} + 1=q(q+1)(q+2)(q+3)+1$$
Let $q=a-frac{3}{2}$
Then
$$S_q=left(a-frac{3}{2}right)left(a-frac{1}{2}right)left(a+frac{1}{2}right)left(a+frac{3}{2}right)+1$$
$$S_q=left(a^2-frac{3^2}{2^2}right)left(a^2-frac{1}{2^2}right)+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{3^2}{4^2}+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{5^2}{4^2}$$
$$S_q=left(a^2-frac{5}{4} right)^2$$
$$S_q=left((q+3/2)^2-frac{5}{4} right)^2$$
$$S_p=left((p-3/2)^2-frac{5}{4} right)^2$$
answered Jan 28 at 15:07
James ArathoonJames Arathoon
1,608423
$endgroup$
Slightly different approach... where $p=q+3$
$$S_q=frac{(p)!}{(p-4)!}+1=frac{(q+3)!}{(q-1)!} + 1=q(q+1)(q+2)(q+3)+1$$
Let $q=a-frac{3}{2}$
Then
$$S_q=left(a-frac{3}{2}right)left(a-frac{1}{2}right)left(a+frac{1}{2}right)left(a+frac{3}{2}right)+1$$
$$S_q=left(a^2-frac{3^2}{2^2}right)left(a^2-frac{1}{2^2}right)+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{3^2}{4^2}+1$$
$$S_q=a^4-frac{10}{4}a^2+frac{5^2}{4^2}$$
$$S_q=left(a^2-frac{5}{4} right)^2$$
$$S_q=left((q+3/2)^2-frac{5}{4} right)^2$$
$$S_p=left((p-3/2)^2-frac{5}{4} right)^2$$
answered Jan 28 at 15:07
James ArathoonJames Arathoon
1,608423
edited Jan 28 at 15:21
answered Jan 28 at 15:07
James ArathoonJames Arathoon
1,608423
answered Jan 28 at 15:07
James ArathoonJames Arathoon
1,608423
answered Jan 28 at 15:07
James ArathoonJames Arathoon
1,608423
1,608423
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I think this is what my teacher expected.
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– Mary Tom
Jan 28 at 15:38
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@MaryTom It then takes just a little bit of thought to know that if an integer is the square of a rational number, it is the square of an integer. Or to expand the expression in the bracket and see that it is an integer. It is not quite immediately obvious that the expression in the final bracket is an integer.
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– Mark Bennet
Jan 28 at 16:51
add a comment |
$begingroup$
I think this is what my teacher expected.
$endgroup$
– Mary Tom
Jan 28 at 15:38
$begingroup$
@MaryTom It then takes just a little bit of thought to know that if an integer is the square of a rational number, it is the square of an integer. Or to expand the expression in the bracket and see that it is an integer. It is not quite immediately obvious that the expression in the final bracket is an integer.
$endgroup$
– Mark Bennet
Jan 28 at 16:51
$begingroup$
I think this is what my teacher expected.
$endgroup$
– Mary Tom
Jan 28 at 15:38
$begingroup$
I think this is what my teacher expected.
$endgroup$
– Mary Tom
Jan 28 at 15:38
$begingroup$
@MaryTom It then takes just a little bit of thought to know that if an integer is the square of a rational number, it is the square of an integer. Or to expand the expression in the bracket and see that it is an integer. It is not quite immediately obvious that the expression in the final bracket is an integer.
$endgroup$
– Mark Bennet
Jan 28 at 16:51
$begingroup$
@MaryTom It then takes just a little bit of thought to know that if an integer is the square of a rational number, it is the square of an integer. Or to expand the expression in the bracket and see that it is an integer. It is not quite immediately obvious that the expression in the final bracket is an integer.
$endgroup$
– Mark Bennet
Jan 28 at 16:51
add a comment |
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$begingroup$
Did you say n when you meant to say p in your title?
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– Mark
Jan 28 at 14:19
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Welcome to MSE! Did you mean for all natural $p$?
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– J. W. Tanner
Jan 28 at 14:19
1
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Are you sure that it is $p(p+1)(p+2)(p+3)+1$ and not $p(p-1)(p-2)(p-3)+1$?
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– user 170039
Jan 28 at 14:20
1
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You are using $p$ and $n$ in confusing ways, Your target seems to be the product of four consecutive integers plus one. But these integers are named inconsistently.
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– Mark Bennet
Jan 28 at 14:24
3
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Your computation appears to be working with $frac {(p+4)!}{p!}+1$ instead of what you wrote. Doesn't change much, but it makes your post unhelpfully confusing.
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– lulu
Jan 28 at 14:24