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
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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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
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New contributor
edited 3 mins ago
J. W. Tanner
956113
956113
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asked 2 hours ago
JayJay
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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., $n = 0, 1, -2, -3$.
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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 $n=1,-3,-2,0$
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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.
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– John Omielan
2 hours ago
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@JohnOmielan Edited.
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– tatan
1 hour ago
add a comment |
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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., $n = 0, 1, -2, -3$.
$endgroup$
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., $n = 0, 1, -2, -3$.
$endgroup$
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., $n = 0, 1, -2, -3$.
$endgroup$
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 2 hours ago
John OmielanJohn Omielan
2,046210
2,046210
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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 $n=1,-3,-2,0$
$endgroup$
$begingroup$
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
2 hours ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
1 hour 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 $n=1,-3,-2,0$
$endgroup$
$begingroup$
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
2 hours ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
1 hour 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 $n=1,-3,-2,0$
$endgroup$
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 $n=1,-3,-2,0$
edited 1 hour ago
answered 2 hours ago
tatantatan
5,70062758
5,70062758
$begingroup$
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
2 hours ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
1 hour 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
2 hours ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
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
2 hours 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
2 hours ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
1 hour ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
1 hour ago
add a comment |
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