Number of solutions in quadratic congruence
I use an example to explain my question:
How many solutions are there to $x^2+3x+18equiv 0$ (mod $28$).
Usually I will encounter these kind of problems. My first step must transform this equation to become 2 equations:
$$x^2+3x+18equiv 0 quad (text{mod }7) $$
$$x^2+3x+18equiv 0quad (text{mod }4) $$
Then, I just plug in the numbers $0$ to $6$ to eq 1 to find out there is one solution $xequiv 2$ mod $7$.
For equation 2, I do the same thing, plug in $0$ to $3$ and there are two solutions:$xequiv 2$ and $xequiv 3$ mod $4$.
Then how should I continue?? Using Chinese Remainder Theorem? Because I have made the original equation into a two simultaneous equations, maybe something like:
$xequiv 2$ mod $7$ and $xequiv 2$ mod $4$
$xequiv 2$ mod $7$ and $xequiv 3$ mod $4$
Is it the general way to do it? But is there a faster way to determine the number of solutions without performing CRT?
number-theory modular-arithmetic
asked Nov 30 at 6:16
Jason Ng
1118
add a comment |
I use an example to explain my question:
How many solutions are there to $x^2+3x+18equiv 0$ (mod $28$).
Usually I will encounter these kind of problems. My first step must transform this equation to become 2 equations:
$$x^2+3x+18equiv 0 quad (text{mod }7) $$
$$x^2+3x+18equiv 0quad (text{mod }4) $$
Then, I just plug in the numbers $0$ to $6$ to eq 1 to find out there is one solution $xequiv 2$ mod $7$.
For equation 2, I do the same thing, plug in $0$ to $3$ and there are two solutions:$xequiv 2$ and $xequiv 3$ mod $4$.
Then how should I continue?? Using Chinese Remainder Theorem? Because I have made the original equation into a two simultaneous equations, maybe something like:
$xequiv 2$ mod $7$ and $xequiv 2$ mod $4$
$xequiv 2$ mod $7$ and $xequiv 3$ mod $4$
Is it the general way to do it? But is there a faster way to determine the number of solutions without performing CRT?
number-theory modular-arithmetic
asked Nov 30 at 6:16
Jason Ng
1118
add a comment |
I use an example to explain my question:
How many solutions are there to $x^2+3x+18equiv 0$ (mod $28$).
Usually I will encounter these kind of problems. My first step must transform this equation to become 2 equations:
$$x^2+3x+18equiv 0 quad (text{mod }7) $$
$$x^2+3x+18equiv 0quad (text{mod }4) $$
Then, I just plug in the numbers $0$ to $6$ to eq 1 to find out there is one solution $xequiv 2$ mod $7$.
For equation 2, I do the same thing, plug in $0$ to $3$ and there are two solutions:$xequiv 2$ and $xequiv 3$ mod $4$.
Then how should I continue?? Using Chinese Remainder Theorem? Because I have made the original equation into a two simultaneous equations, maybe something like:
$xequiv 2$ mod $7$ and $xequiv 2$ mod $4$
$xequiv 2$ mod $7$ and $xequiv 3$ mod $4$
Is it the general way to do it? But is there a faster way to determine the number of solutions without performing CRT?
number-theory modular-arithmetic
asked Nov 30 at 6:16
Jason Ng
1118
I use an example to explain my question:
How many solutions are there to $x^2+3x+18equiv 0$ (mod $28$).
Usually I will encounter these kind of problems. My first step must transform this equation to become 2 equations:
$$x^2+3x+18equiv 0 quad (text{mod }7) $$
$$x^2+3x+18equiv 0quad (text{mod }4) $$
Then, I just plug in the numbers $0$ to $6$ to eq 1 to find out there is one solution $xequiv 2$ mod $7$.
For equation 2, I do the same thing, plug in $0$ to $3$ and there are two solutions:$xequiv 2$ and $xequiv 3$ mod $4$.
Then how should I continue?? Using Chinese Remainder Theorem? Because I have made the original equation into a two simultaneous equations, maybe something like:
$xequiv 2$ mod $7$ and $xequiv 2$ mod $4$
$xequiv 2$ mod $7$ and $xequiv 3$ mod $4$
Is it the general way to do it? But is there a faster way to determine the number of solutions without performing CRT?
number-theory modular-arithmetic
number-theory modular-arithmetic
asked Nov 30 at 6:16
Jason Ng
1118
asked Nov 30 at 6:16
Jason Ng
1118
asked Nov 30 at 6:16
Jason Ng
1118
asked Nov 30 at 6:16
Jason Ng
1118
asked Nov 30 at 6:16
Jason Ng
1118
1118
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Hint. Note that
$$x^2+3x+18equiv x^2+3x-10= (x+5)(x-2)pmod{28}.$$
answered Nov 30 at 6:22
Robert Z
93.2k1061132
Thank you. This is a much better method.
– Jason Ng
Nov 30 at 6:32
Thanks, this is by far the best method
– Alessar
Dec 3 at 12:34
add a comment |
$xequiv2pmod4,xequiv2pmod7implies$lcm$(4,7)|(x-2)implies xequiv2pmod{28}$
For the second, $$7a+2=4b+3iff7a=4b+8-7iffdfrac{7(a+1)}4=b+2$$ which is an integer
$implies4|7(a+1)iff4|(a+1)$ as $(4,7)=1$
$implies a+1=4c$
$implies x=7a+2=7(4c-1)+2=28c-5equiv-5pmod{28}equiv-5+28$
answered Nov 30 at 6:24
lab bhattacharjee
222k15156274
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
});
}
});
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%2f3019723%2fnumber-of-solutions-in-quadratic-congruence%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
Hint. Note that
$$x^2+3x+18equiv x^2+3x-10= (x+5)(x-2)pmod{28}.$$
answered Nov 30 at 6:22
Robert Z
93.2k1061132
Thank you. This is a much better method.
– Jason Ng
Nov 30 at 6:32
Thanks, this is by far the best method
– Alessar
Dec 3 at 12:34
add a comment |
Hint. Note that
$$x^2+3x+18equiv x^2+3x-10= (x+5)(x-2)pmod{28}.$$
answered Nov 30 at 6:22
Robert Z
93.2k1061132
Thank you. This is a much better method.
– Jason Ng
Nov 30 at 6:32
Thanks, this is by far the best method
– Alessar
Dec 3 at 12:34
add a comment |
Hint. Note that
$$x^2+3x+18equiv x^2+3x-10= (x+5)(x-2)pmod{28}.$$
answered Nov 30 at 6:22
Robert Z
93.2k1061132
Hint. Note that
$$x^2+3x+18equiv x^2+3x-10= (x+5)(x-2)pmod{28}.$$
answered Nov 30 at 6:22
Robert Z
93.2k1061132
answered Nov 30 at 6:22
Robert Z
93.2k1061132
answered Nov 30 at 6:22
Robert Z
93.2k1061132
answered Nov 30 at 6:22
Robert Z
93.2k1061132
93.2k1061132
Thank you. This is a much better method.
– Jason Ng
Nov 30 at 6:32
Thanks, this is by far the best method
– Alessar
Dec 3 at 12:34
add a comment |
Thank you. This is a much better method.
– Jason Ng
Nov 30 at 6:32
Thanks, this is by far the best method
– Alessar
Dec 3 at 12:34
Thank you. This is a much better method.
– Jason Ng
Nov 30 at 6:32
Thank you. This is a much better method.
– Jason Ng
Nov 30 at 6:32
Thanks, this is by far the best method
– Alessar
Dec 3 at 12:34
Thanks, this is by far the best method
– Alessar
Dec 3 at 12:34
add a comment |
$xequiv2pmod4,xequiv2pmod7implies$lcm$(4,7)|(x-2)implies xequiv2pmod{28}$
For the second, $$7a+2=4b+3iff7a=4b+8-7iffdfrac{7(a+1)}4=b+2$$ which is an integer
$implies4|7(a+1)iff4|(a+1)$ as $(4,7)=1$
$implies a+1=4c$
$implies x=7a+2=7(4c-1)+2=28c-5equiv-5pmod{28}equiv-5+28$
answered Nov 30 at 6:24
lab bhattacharjee
222k15156274
add a comment |
$xequiv2pmod4,xequiv2pmod7implies$lcm$(4,7)|(x-2)implies xequiv2pmod{28}$
For the second, $$7a+2=4b+3iff7a=4b+8-7iffdfrac{7(a+1)}4=b+2$$ which is an integer
$implies4|7(a+1)iff4|(a+1)$ as $(4,7)=1$
$implies a+1=4c$
$implies x=7a+2=7(4c-1)+2=28c-5equiv-5pmod{28}equiv-5+28$
answered Nov 30 at 6:24
lab bhattacharjee
222k15156274
add a comment |
$xequiv2pmod4,xequiv2pmod7implies$lcm$(4,7)|(x-2)implies xequiv2pmod{28}$
For the second, $$7a+2=4b+3iff7a=4b+8-7iffdfrac{7(a+1)}4=b+2$$ which is an integer
$implies4|7(a+1)iff4|(a+1)$ as $(4,7)=1$
$implies a+1=4c$
$implies x=7a+2=7(4c-1)+2=28c-5equiv-5pmod{28}equiv-5+28$
answered Nov 30 at 6:24
lab bhattacharjee
222k15156274
$xequiv2pmod4,xequiv2pmod7implies$lcm$(4,7)|(x-2)implies xequiv2pmod{28}$
For the second, $$7a+2=4b+3iff7a=4b+8-7iffdfrac{7(a+1)}4=b+2$$ which is an integer
$implies4|7(a+1)iff4|(a+1)$ as $(4,7)=1$
$implies a+1=4c$
$implies x=7a+2=7(4c-1)+2=28c-5equiv-5pmod{28}equiv-5+28$
answered Nov 30 at 6:24
lab bhattacharjee
222k15156274
answered Nov 30 at 6:24
lab bhattacharjee
222k15156274
answered Nov 30 at 6:24
lab bhattacharjee
222k15156274
answered Nov 30 at 6:24
lab bhattacharjee
222k15156274
222k15156274
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3019723%2fnumber-of-solutions-in-quadratic-congruence%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
