Find all solutions when two functions are the same (same result)
2
$begingroup$
Can anybody help me find solutions for $x$ when: $$y_1 = y_2$$
and equations are (for example):
$$y_1 = frac{(x^2+x)}{2} ;; ;y_2 = frac{(x^2+19x-12)}{2}$$
By solving it "brute force" I can tell some of the solutions are:
- $$y_1(5)=y_2(2)=15$$
- $$y_1(15)=y_2(9)=120$$
- $$? ? ?$$
Thank you!
functions systems-of-equations
edited Dec 28 '18 at 10:01
Jaideep Khare
17.8k32669
asked Dec 28 '18 at 9:57
JovanJovan
164
$endgroup$
add a comment |
2
$begingroup$
Can anybody help me find solutions for $x$ when: $$y_1 = y_2$$
and equations are (for example):
$$y_1 = frac{(x^2+x)}{2} ;; ;y_2 = frac{(x^2+19x-12)}{2}$$
By solving it "brute force" I can tell some of the solutions are:
- $$y_1(5)=y_2(2)=15$$
- $$y_1(15)=y_2(9)=120$$
- $$? ? ?$$
Thank you!
functions systems-of-equations
edited Dec 28 '18 at 10:01
Jaideep Khare
17.8k32669
asked Dec 28 '18 at 9:57
JovanJovan
164
$endgroup$
$begingroup$
So, to be clear, when $y_1$ and $y_2$ have the same output, even if $x$ might not be the same... An interesting question. Experience might suggest a clever parameterization, if not brute force, but this is just a guess. Nonetheless, good luck.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:00
$begingroup$
In short, you want to solve diophantine equations (for two variables). See this . It might help a little.
$endgroup$
– Jaideep Khare
Dec 28 '18 at 10:04
$begingroup$
Wouldn't Diophantine equations only suffice for integer solutions? What about real values of $x$, which I feel is more in line with what OP wants.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:05
$begingroup$
There many confusing things here. With usual interpretation we have $y_1=y_2$ iff $x=frac 2 3$ but you are allowing different values of $x$ on LHS and RHS. Also it is not clear if you are looking for integer solutions. There is a continuum of solutions if you don't restrict to integer values of $x$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 10:06
$begingroup$
Hi Kavi, Yes I am looking of course integer solution, sorry for not mentioning it.
$endgroup$
– Jovan
Dec 28 '18 at 10:11
add a comment |
2
2
2
0
$begingroup$
Can anybody help me find solutions for $x$ when: $$y_1 = y_2$$
and equations are (for example):
$$y_1 = frac{(x^2+x)}{2} ;; ;y_2 = frac{(x^2+19x-12)}{2}$$
By solving it "brute force" I can tell some of the solutions are:
- $$y_1(5)=y_2(2)=15$$
- $$y_1(15)=y_2(9)=120$$
- $$? ? ?$$
Thank you!
functions systems-of-equations
edited Dec 28 '18 at 10:01
Jaideep Khare
17.8k32669
asked Dec 28 '18 at 9:57
JovanJovan
164
$endgroup$
Can anybody help me find solutions for $x$ when: $$y_1 = y_2$$
and equations are (for example):
$$y_1 = frac{(x^2+x)}{2} ;; ;y_2 = frac{(x^2+19x-12)}{2}$$
By solving it "brute force" I can tell some of the solutions are:
- $$y_1(5)=y_2(2)=15$$
- $$y_1(15)=y_2(9)=120$$
- $$? ? ?$$
Thank you!
functions systems-of-equations
functions systems-of-equations
edited Dec 28 '18 at 10:01
Jaideep Khare
17.8k32669
asked Dec 28 '18 at 9:57
JovanJovan
164
edited Dec 28 '18 at 10:01
Jaideep Khare
17.8k32669
asked Dec 28 '18 at 9:57
JovanJovan
164
edited Dec 28 '18 at 10:01
Jaideep Khare
17.8k32669
edited Dec 28 '18 at 10:01
Jaideep Khare
17.8k32669
edited Dec 28 '18 at 10:01
Jaideep Khare
17.8k32669
17.8k32669
asked Dec 28 '18 at 9:57
JovanJovan
164
asked Dec 28 '18 at 9:57
JovanJovan
164
asked Dec 28 '18 at 9:57
JovanJovan
164
164
$begingroup$
So, to be clear, when $y_1$ and $y_2$ have the same output, even if $x$ might not be the same... An interesting question. Experience might suggest a clever parameterization, if not brute force, but this is just a guess. Nonetheless, good luck.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:00
$begingroup$
In short, you want to solve diophantine equations (for two variables). See this . It might help a little.
$endgroup$
– Jaideep Khare
Dec 28 '18 at 10:04
$begingroup$
Wouldn't Diophantine equations only suffice for integer solutions? What about real values of $x$, which I feel is more in line with what OP wants.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:05
$begingroup$
There many confusing things here. With usual interpretation we have $y_1=y_2$ iff $x=frac 2 3$ but you are allowing different values of $x$ on LHS and RHS. Also it is not clear if you are looking for integer solutions. There is a continuum of solutions if you don't restrict to integer values of $x$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 10:06
$begingroup$
Hi Kavi, Yes I am looking of course integer solution, sorry for not mentioning it.
$endgroup$
– Jovan
Dec 28 '18 at 10:11
add a comment |
$begingroup$
So, to be clear, when $y_1$ and $y_2$ have the same output, even if $x$ might not be the same... An interesting question. Experience might suggest a clever parameterization, if not brute force, but this is just a guess. Nonetheless, good luck.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:00
$begingroup$
In short, you want to solve diophantine equations (for two variables). See this . It might help a little.
$endgroup$
– Jaideep Khare
Dec 28 '18 at 10:04
$begingroup$
Wouldn't Diophantine equations only suffice for integer solutions? What about real values of $x$, which I feel is more in line with what OP wants.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:05
$begingroup$
There many confusing things here. With usual interpretation we have $y_1=y_2$ iff $x=frac 2 3$ but you are allowing different values of $x$ on LHS and RHS. Also it is not clear if you are looking for integer solutions. There is a continuum of solutions if you don't restrict to integer values of $x$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 10:06
$begingroup$
Hi Kavi, Yes I am looking of course integer solution, sorry for not mentioning it.
$endgroup$
– Jovan
Dec 28 '18 at 10:11
$begingroup$
So, to be clear, when $y_1$ and $y_2$ have the same output, even if $x$ might not be the same... An interesting question. Experience might suggest a clever parameterization, if not brute force, but this is just a guess. Nonetheless, good luck.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:00
$begingroup$
So, to be clear, when $y_1$ and $y_2$ have the same output, even if $x$ might not be the same... An interesting question. Experience might suggest a clever parameterization, if not brute force, but this is just a guess. Nonetheless, good luck.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:00
$begingroup$
In short, you want to solve diophantine equations (for two variables). See this . It might help a little.
$endgroup$
– Jaideep Khare
Dec 28 '18 at 10:04
$begingroup$
In short, you want to solve diophantine equations (for two variables). See this . It might help a little.
$endgroup$
– Jaideep Khare
Dec 28 '18 at 10:04
$begingroup$
Wouldn't Diophantine equations only suffice for integer solutions? What about real values of $x$, which I feel is more in line with what OP wants.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:05
$begingroup$
Wouldn't Diophantine equations only suffice for integer solutions? What about real values of $x$, which I feel is more in line with what OP wants.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:05
$begingroup$
There many confusing things here. With usual interpretation we have $y_1=y_2$ iff $x=frac 2 3$ but you are allowing different values of $x$ on LHS and RHS. Also it is not clear if you are looking for integer solutions. There is a continuum of solutions if you don't restrict to integer values of $x$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 10:06
$begingroup$
There many confusing things here. With usual interpretation we have $y_1=y_2$ iff $x=frac 2 3$ but you are allowing different values of $x$ on LHS and RHS. Also it is not clear if you are looking for integer solutions. There is a continuum of solutions if you don't restrict to integer values of $x$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 10:06
$begingroup$
Hi Kavi, Yes I am looking of course integer solution, sorry for not mentioning it.
$endgroup$
– Jovan
Dec 28 '18 at 10:11
$begingroup$
Hi Kavi, Yes I am looking of course integer solution, sorry for not mentioning it.
$endgroup$
– Jovan
Dec 28 '18 at 10:11
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
HINT:
For integers $a,b$, we wish to solve $$frac{a^2+a}2=frac{b^2+19b-12}2impliesleft(a+frac12right)^2-frac14=left(b+frac{19}2right)^2-frac{361}4-12$$ so $$left(b+frac{19}2right)^2-left(a+frac12right)^2=102implies (a+b+10)(b-a+9)=2cdot3cdot17$$ from which only eight possible combinations exist.
answered Dec 28 '18 at 11:06
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
$begingroup$
tnx for the hint, but can you please drop some more details, so I can actually conclude for next similar problem? Also as what I can see this would also require checking for each of them, or I am missing something?
$endgroup$
– Jovan
Dec 28 '18 at 11:42
$begingroup$
@Jovan The cases are $(a+b+10,b+a+9)=(1,102),(2,51),(3,34),(6,17),(17,6),(34,3),(51,2),(102,1)$. You'll need to solve all of them.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:03
$begingroup$
thank you for your comments, how come that two example pairs are (5,2), (15,9) not in the list?
$endgroup$
– Jovan
Dec 28 '18 at 13:22
$begingroup$
@Jovan Because $(2,5)$ corresponds to $(a,b)$ not $(a+b+10,b-a+9)$ (minus sign typo in comment above). You will need to solve these simultaneous equations - albeit not difficult.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:24
$begingroup$
TNX! Based on which theorem you reconstruct equations and than you got these pairs to be solved? There is no possibility to derive some general "formulation" which excludes solving all this? I mean there is no other solution, which excludes testing various cases?
$endgroup$
– Jovan
Dec 28 '18 at 13:56
|
show 1 more 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
});
}
});
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%2fmath.stackexchange.com%2fquestions%2f3054739%2ffind-all-solutions-when-two-functions-are-the-same-same-result%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
1
$begingroup$
HINT:
For integers $a,b$, we wish to solve $$frac{a^2+a}2=frac{b^2+19b-12}2impliesleft(a+frac12right)^2-frac14=left(b+frac{19}2right)^2-frac{361}4-12$$ so $$left(b+frac{19}2right)^2-left(a+frac12right)^2=102implies (a+b+10)(b-a+9)=2cdot3cdot17$$ from which only eight possible combinations exist.
answered Dec 28 '18 at 11:06
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
$begingroup$
tnx for the hint, but can you please drop some more details, so I can actually conclude for next similar problem? Also as what I can see this would also require checking for each of them, or I am missing something?
$endgroup$
– Jovan
Dec 28 '18 at 11:42
$begingroup$
@Jovan The cases are $(a+b+10,b+a+9)=(1,102),(2,51),(3,34),(6,17),(17,6),(34,3),(51,2),(102,1)$. You'll need to solve all of them.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:03
$begingroup$
thank you for your comments, how come that two example pairs are (5,2), (15,9) not in the list?
$endgroup$
– Jovan
Dec 28 '18 at 13:22
$begingroup$
@Jovan Because $(2,5)$ corresponds to $(a,b)$ not $(a+b+10,b-a+9)$ (minus sign typo in comment above). You will need to solve these simultaneous equations - albeit not difficult.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:24
$begingroup$
TNX! Based on which theorem you reconstruct equations and than you got these pairs to be solved? There is no possibility to derive some general "formulation" which excludes solving all this? I mean there is no other solution, which excludes testing various cases?
$endgroup$
– Jovan
Dec 28 '18 at 13:56
|
show 1 more comment
1
$begingroup$
HINT:
For integers $a,b$, we wish to solve $$frac{a^2+a}2=frac{b^2+19b-12}2impliesleft(a+frac12right)^2-frac14=left(b+frac{19}2right)^2-frac{361}4-12$$ so $$left(b+frac{19}2right)^2-left(a+frac12right)^2=102implies (a+b+10)(b-a+9)=2cdot3cdot17$$ from which only eight possible combinations exist.
answered Dec 28 '18 at 11:06
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
$begingroup$
tnx for the hint, but can you please drop some more details, so I can actually conclude for next similar problem? Also as what I can see this would also require checking for each of them, or I am missing something?
$endgroup$
– Jovan
Dec 28 '18 at 11:42
$begingroup$
@Jovan The cases are $(a+b+10,b+a+9)=(1,102),(2,51),(3,34),(6,17),(17,6),(34,3),(51,2),(102,1)$. You'll need to solve all of them.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:03
$begingroup$
thank you for your comments, how come that two example pairs are (5,2), (15,9) not in the list?
$endgroup$
– Jovan
Dec 28 '18 at 13:22
$begingroup$
@Jovan Because $(2,5)$ corresponds to $(a,b)$ not $(a+b+10,b-a+9)$ (minus sign typo in comment above). You will need to solve these simultaneous equations - albeit not difficult.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:24
$begingroup$
TNX! Based on which theorem you reconstruct equations and than you got these pairs to be solved? There is no possibility to derive some general "formulation" which excludes solving all this? I mean there is no other solution, which excludes testing various cases?
$endgroup$
– Jovan
Dec 28 '18 at 13:56
|
show 1 more comment
1
1
1
$begingroup$
HINT:
For integers $a,b$, we wish to solve $$frac{a^2+a}2=frac{b^2+19b-12}2impliesleft(a+frac12right)^2-frac14=left(b+frac{19}2right)^2-frac{361}4-12$$ so $$left(b+frac{19}2right)^2-left(a+frac12right)^2=102implies (a+b+10)(b-a+9)=2cdot3cdot17$$ from which only eight possible combinations exist.
answered Dec 28 '18 at 11:06
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
HINT:
For integers $a,b$, we wish to solve $$frac{a^2+a}2=frac{b^2+19b-12}2impliesleft(a+frac12right)^2-frac14=left(b+frac{19}2right)^2-frac{361}4-12$$ so $$left(b+frac{19}2right)^2-left(a+frac12right)^2=102implies (a+b+10)(b-a+9)=2cdot3cdot17$$ from which only eight possible combinations exist.
answered Dec 28 '18 at 11:06
TheSimpliFireTheSimpliFire
13.2k62464
answered Dec 28 '18 at 11:06
TheSimpliFireTheSimpliFire
13.2k62464
answered Dec 28 '18 at 11:06
TheSimpliFireTheSimpliFire
13.2k62464
answered Dec 28 '18 at 11:06
TheSimpliFireTheSimpliFire
13.2k62464
13.2k62464
$begingroup$
tnx for the hint, but can you please drop some more details, so I can actually conclude for next similar problem? Also as what I can see this would also require checking for each of them, or I am missing something?
$endgroup$
– Jovan
Dec 28 '18 at 11:42
$begingroup$
@Jovan The cases are $(a+b+10,b+a+9)=(1,102),(2,51),(3,34),(6,17),(17,6),(34,3),(51,2),(102,1)$. You'll need to solve all of them.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:03
$begingroup$
thank you for your comments, how come that two example pairs are (5,2), (15,9) not in the list?
$endgroup$
– Jovan
Dec 28 '18 at 13:22
$begingroup$
@Jovan Because $(2,5)$ corresponds to $(a,b)$ not $(a+b+10,b-a+9)$ (minus sign typo in comment above). You will need to solve these simultaneous equations - albeit not difficult.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:24
$begingroup$
TNX! Based on which theorem you reconstruct equations and than you got these pairs to be solved? There is no possibility to derive some general "formulation" which excludes solving all this? I mean there is no other solution, which excludes testing various cases?
$endgroup$
– Jovan
Dec 28 '18 at 13:56
|
show 1 more comment
$begingroup$
tnx for the hint, but can you please drop some more details, so I can actually conclude for next similar problem? Also as what I can see this would also require checking for each of them, or I am missing something?
$endgroup$
– Jovan
Dec 28 '18 at 11:42
$begingroup$
@Jovan The cases are $(a+b+10,b+a+9)=(1,102),(2,51),(3,34),(6,17),(17,6),(34,3),(51,2),(102,1)$. You'll need to solve all of them.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:03
$begingroup$
thank you for your comments, how come that two example pairs are (5,2), (15,9) not in the list?
$endgroup$
– Jovan
Dec 28 '18 at 13:22
$begingroup$
@Jovan Because $(2,5)$ corresponds to $(a,b)$ not $(a+b+10,b-a+9)$ (minus sign typo in comment above). You will need to solve these simultaneous equations - albeit not difficult.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:24
$begingroup$
TNX! Based on which theorem you reconstruct equations and than you got these pairs to be solved? There is no possibility to derive some general "formulation" which excludes solving all this? I mean there is no other solution, which excludes testing various cases?
$endgroup$
– Jovan
Dec 28 '18 at 13:56
$begingroup$
tnx for the hint, but can you please drop some more details, so I can actually conclude for next similar problem? Also as what I can see this would also require checking for each of them, or I am missing something?
$endgroup$
– Jovan
Dec 28 '18 at 11:42
$begingroup$
tnx for the hint, but can you please drop some more details, so I can actually conclude for next similar problem? Also as what I can see this would also require checking for each of them, or I am missing something?
$endgroup$
– Jovan
Dec 28 '18 at 11:42
$begingroup$
@Jovan The cases are $(a+b+10,b+a+9)=(1,102),(2,51),(3,34),(6,17),(17,6),(34,3),(51,2),(102,1)$. You'll need to solve all of them.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:03
$begingroup$
@Jovan The cases are $(a+b+10,b+a+9)=(1,102),(2,51),(3,34),(6,17),(17,6),(34,3),(51,2),(102,1)$. You'll need to solve all of them.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:03
$begingroup$
thank you for your comments, how come that two example pairs are (5,2), (15,9) not in the list?
$endgroup$
– Jovan
Dec 28 '18 at 13:22
$begingroup$
thank you for your comments, how come that two example pairs are (5,2), (15,9) not in the list?
$endgroup$
– Jovan
Dec 28 '18 at 13:22
$begingroup$
@Jovan Because $(2,5)$ corresponds to $(a,b)$ not $(a+b+10,b-a+9)$ (minus sign typo in comment above). You will need to solve these simultaneous equations - albeit not difficult.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:24
$begingroup$
@Jovan Because $(2,5)$ corresponds to $(a,b)$ not $(a+b+10,b-a+9)$ (minus sign typo in comment above). You will need to solve these simultaneous equations - albeit not difficult.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:24
$begingroup$
TNX! Based on which theorem you reconstruct equations and than you got these pairs to be solved? There is no possibility to derive some general "formulation" which excludes solving all this? I mean there is no other solution, which excludes testing various cases?
$endgroup$
– Jovan
Dec 28 '18 at 13:56
$begingroup$
TNX! Based on which theorem you reconstruct equations and than you got these pairs to be solved? There is no possibility to derive some general "formulation" which excludes solving all this? I mean there is no other solution, which excludes testing various cases?
$endgroup$
– Jovan
Dec 28 '18 at 13:56
|
show 1 more comment
draft saved
draft discarded
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.
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%2fmath.stackexchange.com%2fquestions%2f3054739%2ffind-all-solutions-when-two-functions-are-the-same-same-result%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
$begingroup$
So, to be clear, when $y_1$ and $y_2$ have the same output, even if $x$ might not be the same... An interesting question. Experience might suggest a clever parameterization, if not brute force, but this is just a guess. Nonetheless, good luck.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:00
$begingroup$
In short, you want to solve diophantine equations (for two variables). See this . It might help a little.
$endgroup$
– Jaideep Khare
Dec 28 '18 at 10:04
$begingroup$
Wouldn't Diophantine equations only suffice for integer solutions? What about real values of $x$, which I feel is more in line with what OP wants.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 10:05
$begingroup$
There many confusing things here. With usual interpretation we have $y_1=y_2$ iff $x=frac 2 3$ but you are allowing different values of $x$ on LHS and RHS. Also it is not clear if you are looking for integer solutions. There is a continuum of solutions if you don't restrict to integer values of $x$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 10:06
$begingroup$
Hi Kavi, Yes I am looking of course integer solution, sorry for not mentioning it.
$endgroup$
– Jovan
Dec 28 '18 at 10:11