Solution of Diophantine Equation $1+a^6=x^2$
Multi tool use
$begingroup$
Does the equation $1+a^6=x^2$ have any other integer solutions except the trivial one $a=0,x=1$?
number-theory diophantine-equations
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
$endgroup$
add a comment |
$begingroup$
Does the equation $1+a^6=x^2$ have any other integer solutions except the trivial one $a=0,x=1$?
number-theory diophantine-equations
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
$endgroup$
add a comment |
$begingroup$
Does the equation $1+a^6=x^2$ have any other integer solutions except the trivial one $a=0,x=1$?
number-theory diophantine-equations
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
$endgroup$
Does the equation $1+a^6=x^2$ have any other integer solutions except the trivial one $a=0,x=1$?
number-theory diophantine-equations
number-theory diophantine-equations
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
edited Dec 31 '18 at 7:56
dmtri
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
edited Dec 31 '18 at 7:56
dmtri
1,7652521
edited Dec 31 '18 at 7:56
dmtri
1,7652521
edited Dec 31 '18 at 7:56
dmtri
1,7652521
1,7652521
asked Dec 31 '18 at 6:32
ershersh
603113
asked Dec 31 '18 at 6:32
ershersh
603113
asked Dec 31 '18 at 6:32
ershersh
603113
603113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
Your Answer
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
});
}
});
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%2f3057481%2fsolution-of-diophantine-equation-1a6-x2%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
$begingroup$
It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
It's $$(x-a^3)(x+a^3)=1$$ and solve two systems:
$$x-a^3=1$$ and $$x+a^3=1$$ or
$$x-a^3=-1$$ and $$x+a^3=-1,$$
which gives the answer for $(x,a)$:
$${(1,0), (-1,0)}.$$
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
edited Dec 31 '18 at 7:30
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
answered Dec 31 '18 at 7:21
Michael RozenbergMichael Rozenberg
111k1897201
111k1897201
add a comment |
add a comment |
$begingroup$
$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
$begingroup$
$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
$begingroup$
$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
$endgroup$
$x^2$ and $a^6$ are both perfect squares. And the only two perfect squares that differ by $1$ are $0$ and $1$ (because for any $nge 0$, the next biggest square after $n^2$ is $n^2+2n+1$).
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
answered Dec 31 '18 at 7:33
TonyKTonyK
44.1k358137
44.1k358137
add a comment |
add a comment |
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%2f3057481%2fsolution-of-diophantine-equation-1a6-x2%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
RSbSA7 kFEhR,GEOgASsPfhre9,ruHB4d2i LGLA CLU hz INUkOp 1V4eGo,esE sCzV
