Mixing
How to find an integer solution?
2
$begingroup$
Given:
3*(c^3 - c^2*b - c*a^2 + a^2*b) ==
a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - 3 a b c
Need to solve for all three variables as integers. Problem is to find one solution where all variables are positive.
equation-solving
asked Jan 27 at 9:39
DaleDale
1305
$endgroup$
add a comment |
2
$begingroup$
Given:
3*(c^3 - c^2*b - c*a^2 + a^2*b) ==
a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - 3 a b c
Need to solve for all three variables as integers. Problem is to find one solution where all variables are positive.
equation-solving
asked Jan 27 at 9:39
DaleDale
1305
$endgroup$
$begingroup$
please review my edit. Some of your formatting of the formula made it unclear and I'm not sure that I got it completely right.
$endgroup$
– Carl Lange
Jan 27 at 9:56
1
$begingroup$
@carl lange Gostei muito do seu canal no YouTube
$endgroup$
– LCarvalho
Jan 27 at 11:19
$begingroup$
@LCarvalho Obrigado pelas suas palavras gentis!!
$endgroup$
– Carl Lange
Jan 27 at 15:14
add a comment |
2
2
2
$begingroup$
Given:
3*(c^3 - c^2*b - c*a^2 + a^2*b) ==
a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - 3 a b c
Need to solve for all three variables as integers. Problem is to find one solution where all variables are positive.
equation-solving
asked Jan 27 at 9:39
DaleDale
1305
$endgroup$
Given:
3*(c^3 - c^2*b - c*a^2 + a^2*b) ==
a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - 3 a b c
Need to solve for all three variables as integers. Problem is to find one solution where all variables are positive.
equation-solving
equation-solving
asked Jan 27 at 9:39
DaleDale
1305
asked Jan 27 at 9:39
DaleDale
1305
edited Jan 27 at 14:09
Dale
asked Jan 27 at 9:39
DaleDale
1305
asked Jan 27 at 9:39
DaleDale
1305
asked Jan 27 at 9:39
DaleDale
1305
1305
$begingroup$
please review my edit. Some of your formatting of the formula made it unclear and I'm not sure that I got it completely right.
$endgroup$
– Carl Lange
Jan 27 at 9:56
1
$begingroup$
@carl lange Gostei muito do seu canal no YouTube
$endgroup$
– LCarvalho
Jan 27 at 11:19
$begingroup$
@LCarvalho Obrigado pelas suas palavras gentis!!
$endgroup$
– Carl Lange
Jan 27 at 15:14
add a comment |
$begingroup$
please review my edit. Some of your formatting of the formula made it unclear and I'm not sure that I got it completely right.
$endgroup$
– Carl Lange
Jan 27 at 9:56
1
$begingroup$
@carl lange Gostei muito do seu canal no YouTube
$endgroup$
– LCarvalho
Jan 27 at 11:19
$begingroup$
@LCarvalho Obrigado pelas suas palavras gentis!!
$endgroup$
– Carl Lange
Jan 27 at 15:14
$begingroup$
please review my edit. Some of your formatting of the formula made it unclear and I'm not sure that I got it completely right.
$endgroup$
– Carl Lange
Jan 27 at 9:56
$begingroup$
please review my edit. Some of your formatting of the formula made it unclear and I'm not sure that I got it completely right.
$endgroup$
– Carl Lange
Jan 27 at 9:56
1
1
$begingroup$
@carl lange Gostei muito do seu canal no YouTube
$endgroup$
– LCarvalho
Jan 27 at 11:19
$begingroup$
@carl lange Gostei muito do seu canal no YouTube
$endgroup$
– LCarvalho
Jan 27 at 11:19
$begingroup$
@LCarvalho Obrigado pelas suas palavras gentis!!
$endgroup$
– Carl Lange
Jan 27 at 15:14
$begingroup$
@LCarvalho Obrigado pelas suas palavras gentis!!
$endgroup$
– Carl Lange
Jan 27 at 15:14
add a comment |
1 Answer
1
active
oldest
votes
7
$begingroup$
Use FindInstance.
f = 3*(c^3 - c^2 b - c a^2 + a^2 b) ==
a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - (3 a b c)
FindInstance[f, {a, b, c}, Integers, 3]
{{a -> -130, b -> -130, c -> 65}, {a -> -1, b -> 0, c -> 1}, {a -> 1,
b -> 0, c -> -1}}
We can test the solutions out:
f /. {a -> -130, b -> -130, c -> 65}
True
answered Jan 27 at 10:03
Carl LangeCarl Lange
5,02711141
$endgroup$
$begingroup$
Nice work! Do you think it would be possible to have all variables positive? I meant to add this to the question/will edit.
$endgroup$
– Dale
Jan 27 at 14:08
$begingroup$
Found this: mathematica.stackexchange.com/questions/74215/…
$endgroup$
– Dale
Jan 27 at 14:14
1
$begingroup$
@Dale Yes, simply add the extra constraints to the first argument inFindInstance:FindInstance[{f, a > 0, b > 0, c > 0}, {a, b, c}, Integers, 1]However, in this case it does not appear that a solution exists.
$endgroup$
– Carl Lange
Jan 27 at 15:13
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: "387"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
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%2fmathematica.stackexchange.com%2fquestions%2f190323%2fhow-to-find-an-integer-solution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
Use FindInstance.
f = 3*(c^3 - c^2 b - c a^2 + a^2 b) ==
a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - (3 a b c)
FindInstance[f, {a, b, c}, Integers, 3]
{{a -> -130, b -> -130, c -> 65}, {a -> -1, b -> 0, c -> 1}, {a -> 1,
b -> 0, c -> -1}}
We can test the solutions out:
f /. {a -> -130, b -> -130, c -> 65}
True
answered Jan 27 at 10:03
Carl LangeCarl Lange
5,02711141
$endgroup$
$begingroup$
Nice work! Do you think it would be possible to have all variables positive? I meant to add this to the question/will edit.
$endgroup$
– Dale
Jan 27 at 14:08
$begingroup$
Found this: mathematica.stackexchange.com/questions/74215/…
$endgroup$
– Dale
Jan 27 at 14:14
1
$begingroup$
@Dale Yes, simply add the extra constraints to the first argument inFindInstance:FindInstance[{f, a > 0, b > 0, c > 0}, {a, b, c}, Integers, 1]However, in this case it does not appear that a solution exists.
$endgroup$
– Carl Lange
Jan 27 at 15:13
add a comment |
7
$begingroup$
Use FindInstance.
f = 3*(c^3 - c^2 b - c a^2 + a^2 b) ==
a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - (3 a b c)
FindInstance[f, {a, b, c}, Integers, 3]
{{a -> -130, b -> -130, c -> 65}, {a -> -1, b -> 0, c -> 1}, {a -> 1,
b -> 0, c -> -1}}
We can test the solutions out:
f /. {a -> -130, b -> -130, c -> 65}
True
answered Jan 27 at 10:03
Carl LangeCarl Lange
5,02711141
$endgroup$
$begingroup$
Nice work! Do you think it would be possible to have all variables positive? I meant to add this to the question/will edit.
$endgroup$
– Dale
Jan 27 at 14:08
$begingroup$
Found this: mathematica.stackexchange.com/questions/74215/…
$endgroup$
– Dale
Jan 27 at 14:14
1
$begingroup$
@Dale Yes, simply add the extra constraints to the first argument inFindInstance:FindInstance[{f, a > 0, b > 0, c > 0}, {a, b, c}, Integers, 1]However, in this case it does not appear that a solution exists.
$endgroup$
– Carl Lange
Jan 27 at 15:13
add a comment |
7
7
7
$begingroup$
Use FindInstance.
f = 3*(c^3 - c^2 b - c a^2 + a^2 b) ==
a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - (3 a b c)
FindInstance[f, {a, b, c}, Integers, 3]
{{a -> -130, b -> -130, c -> 65}, {a -> -1, b -> 0, c -> 1}, {a -> 1,
b -> 0, c -> -1}}
We can test the solutions out:
f /. {a -> -130, b -> -130, c -> 65}
True
answered Jan 27 at 10:03
Carl LangeCarl Lange
5,02711141
$endgroup$
Use FindInstance.
f = 3*(c^3 - c^2 b - c a^2 + a^2 b) ==
a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - (3 a b c)
FindInstance[f, {a, b, c}, Integers, 3]
{{a -> -130, b -> -130, c -> 65}, {a -> -1, b -> 0, c -> 1}, {a -> 1,
b -> 0, c -> -1}}
We can test the solutions out:
f /. {a -> -130, b -> -130, c -> 65}
True
answered Jan 27 at 10:03
Carl LangeCarl Lange
5,02711141
answered Jan 27 at 10:03
Carl LangeCarl Lange
5,02711141
answered Jan 27 at 10:03
Carl LangeCarl Lange
5,02711141
answered Jan 27 at 10:03
Carl LangeCarl Lange
5,02711141
5,02711141
$begingroup$
Nice work! Do you think it would be possible to have all variables positive? I meant to add this to the question/will edit.
$endgroup$
– Dale
Jan 27 at 14:08
$begingroup$
Found this: mathematica.stackexchange.com/questions/74215/…
$endgroup$
– Dale
Jan 27 at 14:14
1
$begingroup$
@Dale Yes, simply add the extra constraints to the first argument inFindInstance:FindInstance[{f, a > 0, b > 0, c > 0}, {a, b, c}, Integers, 1]However, in this case it does not appear that a solution exists.
$endgroup$
– Carl Lange
Jan 27 at 15:13
add a comment |
$begingroup$
Nice work! Do you think it would be possible to have all variables positive? I meant to add this to the question/will edit.
$endgroup$
– Dale
Jan 27 at 14:08
$begingroup$
Found this: mathematica.stackexchange.com/questions/74215/…
$endgroup$
– Dale
Jan 27 at 14:14
1
$begingroup$
@Dale Yes, simply add the extra constraints to the first argument inFindInstance:FindInstance[{f, a > 0, b > 0, c > 0}, {a, b, c}, Integers, 1]However, in this case it does not appear that a solution exists.
$endgroup$
– Carl Lange
Jan 27 at 15:13
$begingroup$
Nice work! Do you think it would be possible to have all variables positive? I meant to add this to the question/will edit.
$endgroup$
– Dale
Jan 27 at 14:08
$begingroup$
Nice work! Do you think it would be possible to have all variables positive? I meant to add this to the question/will edit.
$endgroup$
– Dale
Jan 27 at 14:08
$begingroup$
Found this: mathematica.stackexchange.com/questions/74215/…
$endgroup$
– Dale
Jan 27 at 14:14
$begingroup$
Found this: mathematica.stackexchange.com/questions/74215/…
$endgroup$
– Dale
Jan 27 at 14:14
1
1
$begingroup$
@Dale Yes, simply add the extra constraints to the first argument in
FindInstance: FindInstance[{f, a > 0, b > 0, c > 0}, {a, b, c}, Integers, 1] However, in this case it does not appear that a solution exists.$endgroup$
– Carl Lange
Jan 27 at 15:13
$begingroup$
@Dale Yes, simply add the extra constraints to the first argument in
FindInstance: FindInstance[{f, a > 0, b > 0, c > 0}, {a, b, c}, Integers, 1] However, in this case it does not appear that a solution exists.$endgroup$
– Carl Lange
Jan 27 at 15:13
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Mathematica 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.
draft saved
draft discarded
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%2fmathematica.stackexchange.com%2fquestions%2f190323%2fhow-to-find-an-integer-solution%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
$begingroup$
please review my edit. Some of your formatting of the formula made it unclear and I'm not sure that I got it completely right.
$endgroup$
– Carl Lange
Jan 27 at 9:56
1
$begingroup$
@carl lange Gostei muito do seu canal no YouTube
$endgroup$
– LCarvalho
Jan 27 at 11:19
$begingroup$
@LCarvalho Obrigado pelas suas palavras gentis!!
$endgroup$
– Carl Lange
Jan 27 at 15:14