Find all integers n (positive, negative, or zero) so that n^3 – 1 is divisible by n + 1
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I've tried to rearrange n^3 - 1 (turning it into (n-1)(n^2+n+1)) and play around with it, but I haven't been able to figure it out. Within the text this problem was in, the chapter taught you divisibility and the Euclidean Algorithm so I'm assuming the answer will utilize those tools.
proof-writing divisibility
asked 1 hour ago
JayJay
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add a comment |
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I've tried to rearrange n^3 - 1 (turning it into (n-1)(n^2+n+1)) and play around with it, but I haven't been able to figure it out. Within the text this problem was in, the chapter taught you divisibility and the Euclidean Algorithm so I'm assuming the answer will utilize those tools.
proof-writing divisibility
asked 1 hour ago
JayJay
61
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I've tried to rearrange n^3 - 1 (turning it into (n-1)(n^2+n+1)) and play around with it, but I haven't been able to figure it out. Within the text this problem was in, the chapter taught you divisibility and the Euclidean Algorithm so I'm assuming the answer will utilize those tools.
proof-writing divisibility
asked 1 hour ago
JayJay
61
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I've tried to rearrange n^3 - 1 (turning it into (n-1)(n^2+n+1)) and play around with it, but I haven't been able to figure it out. Within the text this problem was in, the chapter taught you divisibility and the Euclidean Algorithm so I'm assuming the answer will utilize those tools.
proof-writing divisibility
proof-writing divisibility
asked 1 hour ago
JayJay
61
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
JayJay
61
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
JayJay
61
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
JayJay
61
asked 1 hour ago
JayJay
61
61
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
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$begingroup$
If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.
answered 1 hour ago
John OmielanJohn Omielan
2,046210
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add a comment |
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John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:
$$n^3-1=(n+1)(n^2-n+1)-2$$
Dividing by $n+1$ gives
$$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$
Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$
answered 1 hour ago
tatantatan
5,69062758
$endgroup$
$begingroup$
The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan Edited.
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– tatan
37 mins ago
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.
answered 1 hour ago
John OmielanJohn Omielan
2,046210
$endgroup$
add a comment |
$begingroup$
If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.
answered 1 hour ago
John OmielanJohn Omielan
2,046210
$endgroup$
add a comment |
$begingroup$
If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.
answered 1 hour ago
John OmielanJohn Omielan
2,046210
$endgroup$
If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.
answered 1 hour ago
John OmielanJohn Omielan
2,046210
answered 1 hour ago
John OmielanJohn Omielan
2,046210
answered 1 hour ago
John OmielanJohn Omielan
2,046210
answered 1 hour ago
John OmielanJohn Omielan
2,046210
2,046210
add a comment |
add a comment |
$begingroup$
John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:
$$n^3-1=(n+1)(n^2-n+1)-2$$
Dividing by $n+1$ gives
$$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$
Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$
answered 1 hour ago
tatantatan
5,69062758
$endgroup$
$begingroup$
The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
37 mins ago
add a comment |
$begingroup$
John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:
$$n^3-1=(n+1)(n^2-n+1)-2$$
Dividing by $n+1$ gives
$$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$
Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$
answered 1 hour ago
tatantatan
5,69062758
$endgroup$
$begingroup$
The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
37 mins ago
add a comment |
$begingroup$
John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:
$$n^3-1=(n+1)(n^2-n+1)-2$$
Dividing by $n+1$ gives
$$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$
Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$
answered 1 hour ago
tatantatan
5,69062758
$endgroup$
John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:
$$n^3-1=(n+1)(n^2-n+1)-2$$
Dividing by $n+1$ gives
$$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$
Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$
answered 1 hour ago
tatantatan
5,69062758
edited 37 mins ago
answered 1 hour ago
tatantatan
5,69062758
answered 1 hour ago
tatantatan
5,69062758
answered 1 hour ago
tatantatan
5,69062758
5,69062758
$begingroup$
The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
37 mins ago
add a comment |
$begingroup$
The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
37 mins ago
$begingroup$
The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
37 mins ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
37 mins ago
add a comment |
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