Find all integers n (positive, negative, or zero) so that n^3 – 1 is divisible by n + 1
$begingroup$
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 2 hours 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 2 hours 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 2 hours 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 2 hours 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 2 hours 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 2 hours 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 2 hours ago
JayJay
61
asked 2 hours 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
2
active
oldest
votes
$begingroup$
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 2 hours ago
John OmielanJohn Omielan
2,046210
$endgroup$
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 2 hours ago
tatantatan
5,70062758
$endgroup$
$begingroup$
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
1 hour ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Jay is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3098147%2ffind-all-integers-n-positive-negative-or-zero-so-that-n3-1-is-divisible-b%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 2 hours 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 2 hours 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 2 hours 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 2 hours ago
John OmielanJohn Omielan
2,046210
answered 2 hours ago
John OmielanJohn Omielan
2,046210
answered 2 hours ago
John OmielanJohn Omielan
2,046210
answered 2 hours 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 2 hours ago
tatantatan
5,70062758
$endgroup$
$begingroup$
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
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 $n=1,-3,-2,0$
answered 2 hours ago
tatantatan
5,70062758
$endgroup$
$begingroup$
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
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 $n=1,-3,-2,0$
answered 2 hours ago
tatantatan
5,70062758
$endgroup$
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 2 hours ago
tatantatan
5,70062758
edited 1 hour ago
answered 2 hours ago
tatantatan
5,70062758
answered 2 hours ago
tatantatan
5,70062758
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
1 hour 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
1 hour 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
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
1 hour ago
$begingroup$
@JohnOmielan Edited.
$endgroup$
– tatan
1 hour ago
add a comment |
Jay is a new contributor. Be nice, and check out our Code of Conduct.
Jay is a new contributor. Be nice, and check out our Code of Conduct.
Jay is a new contributor. Be nice, and check out our Code of Conduct.
Jay is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3098147%2ffind-all-integers-n-positive-negative-or-zero-so-that-n3-1-is-divisible-b%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
