Show solution to recurrence is never a squarePrimes of the form 1..1Integers divide several solutions to Greatest Common Divisor equationProving a solution to a double recurrence is exhaustiveNumbers that are clearly NOT a SquareProcedural question regarding number of steps in Euclidean algorithmLimit of sequence in which each term is the average of its preceding k terms.Geometric represntation of terms for reccurence sequence with $f(x)=frac-12 x+3$?How many digits occur in the $100^th$ term of the following recurrence?Convergence of a linear recurrence equationDerivative of integral for non-negative part functions
Are there any sonatas with only two sections?
Why are solar panels kept tilted?
Formal Definition of Dot Product
Did galley captains put corks in the mouths of slave rowers to keep them quiet?
Alias for root of a polynomial
Biology of a Firestarter
Do we have C++20 ranges library in GCC 9?
the grammar about `adv adv` as 'too quickly'
Is there an academic word that means "to split hairs over"?
Why didn't the Avengers use this object earlier?
Adding labels and comments to a matrix
Can multiple outlets be directly attached to a single breaker?
Use of さ as a filler
Single word that parallels "Recent" when discussing the near future
Why did the metro bus stop at each railway crossing, despite no warning indicating a train was coming?
How to cope with regret and shame about not fully utilizing opportunities during PhD?
How might a landlocked lake become a complete ecosystem?
Offered a new position but unknown about salary?
Given 0s on Assignments with suspected and dismissed cheating?
How do I identify the partitions of my hard drive in order to then shred them all?
Is 12 minutes connection in Bristol Temple Meads long enough?
A case where Bishop for knight isn't a good trade
Will a coyote attack my dog on a leash while I'm on a hiking trail?
Promotion comes with unexpected 24/7/365 on-call
Show solution to recurrence is never a square
Primes of the form 1..1Integers divide several solutions to Greatest Common Divisor equationProving a solution to a double recurrence is exhaustiveNumbers that are clearly NOT a SquareProcedural question regarding number of steps in Euclidean algorithmLimit of sequence in which each term is the average of its preceding k terms.Geometric represntation of terms for reccurence sequence with $f(x)=frac-12 x+3$?How many digits occur in the $100^th$ term of the following recurrence?Convergence of a linear recurrence equationDerivative of integral for non-negative part functions
$begingroup$
Cute problem I saw on quora:
If
$u_1 = 1,
u_n+1 = n+sum_k=1^n u_k^2
$,
show that,
for $n ge 2$,
$u_n$ is never a square.
$
n=1: u_2 = 1+1 = 2\
n=2: u_3 = 2+1+4 = 7\
n=3: u_4 = 3+1+4+49 = 57\
$
And, as usual,
I have a generalization:
If
$a ge 1, b ge 0, u_1 = 1,
u_n+1 = an+b+sum_k=1^n u_k^2,
a, b in mathbbZ,
$then
for $n ge 3$,
$u_n$ is never a square.
Note:
My solutions to these
do not involve
any explicit expressions
for the $u_n$.
elementary-number-theory recurrence-relations square-numbers
asked 3 hours ago
marty cohenmarty cohen
76.8k549130
$endgroup$
add a comment |
$begingroup$
Cute problem I saw on quora:
If
$u_1 = 1,
u_n+1 = n+sum_k=1^n u_k^2
$,
show that,
for $n ge 2$,
$u_n$ is never a square.
$
n=1: u_2 = 1+1 = 2\
n=2: u_3 = 2+1+4 = 7\
n=3: u_4 = 3+1+4+49 = 57\
$
And, as usual,
I have a generalization:
If
$a ge 1, b ge 0, u_1 = 1,
u_n+1 = an+b+sum_k=1^n u_k^2,
a, b in mathbbZ,
$then
for $n ge 3$,
$u_n$ is never a square.
Note:
My solutions to these
do not involve
any explicit expressions
for the $u_n$.
elementary-number-theory recurrence-relations square-numbers
asked 3 hours ago
marty cohenmarty cohen
76.8k549130
$endgroup$
1
$begingroup$
I see the answer from considering modulo $5$. I wonder if it could also be proved that $u_n$ is never a square because $u_n-1<sqrtu_n<u_n-1+1$
$endgroup$
– J. W. Tanner
3 hours ago
add a comment |
$begingroup$
Cute problem I saw on quora:
If
$u_1 = 1,
u_n+1 = n+sum_k=1^n u_k^2
$,
show that,
for $n ge 2$,
$u_n$ is never a square.
$
n=1: u_2 = 1+1 = 2\
n=2: u_3 = 2+1+4 = 7\
n=3: u_4 = 3+1+4+49 = 57\
$
And, as usual,
I have a generalization:
If
$a ge 1, b ge 0, u_1 = 1,
u_n+1 = an+b+sum_k=1^n u_k^2,
a, b in mathbbZ,
$then
for $n ge 3$,
$u_n$ is never a square.
Note:
My solutions to these
do not involve
any explicit expressions
for the $u_n$.
elementary-number-theory recurrence-relations square-numbers
asked 3 hours ago
marty cohenmarty cohen
76.8k549130
$endgroup$
Cute problem I saw on quora:
If
$u_1 = 1,
u_n+1 = n+sum_k=1^n u_k^2
$,
show that,
for $n ge 2$,
$u_n$ is never a square.
$
n=1: u_2 = 1+1 = 2\
n=2: u_3 = 2+1+4 = 7\
n=3: u_4 = 3+1+4+49 = 57\
$
And, as usual,
I have a generalization:
If
$a ge 1, b ge 0, u_1 = 1,
u_n+1 = an+b+sum_k=1^n u_k^2,
a, b in mathbbZ,
$then
for $n ge 3$,
$u_n$ is never a square.
Note:
My solutions to these
do not involve
any explicit expressions
for the $u_n$.
elementary-number-theory recurrence-relations square-numbers
elementary-number-theory recurrence-relations square-numbers
asked 3 hours ago
marty cohenmarty cohen
76.8k549130
asked 3 hours ago
marty cohenmarty cohen
76.8k549130
asked 3 hours ago
marty cohenmarty cohen
76.8k549130
asked 3 hours ago
marty cohenmarty cohen
76.8k549130
asked 3 hours ago
marty cohenmarty cohen
76.8k549130
76.8k549130
1
$begingroup$
I see the answer from considering modulo $5$. I wonder if it could also be proved that $u_n$ is never a square because $u_n-1<sqrtu_n<u_n-1+1$
$endgroup$
– J. W. Tanner
3 hours ago
add a comment |
1
$begingroup$
I see the answer from considering modulo $5$. I wonder if it could also be proved that $u_n$ is never a square because $u_n-1<sqrtu_n<u_n-1+1$
$endgroup$
– J. W. Tanner
3 hours ago
1
1
$begingroup$
I see the answer from considering modulo $5$. I wonder if it could also be proved that $u_n$ is never a square because $u_n-1<sqrtu_n<u_n-1+1$
$endgroup$
– J. W. Tanner
3 hours ago
$begingroup$
I see the answer from considering modulo $5$. I wonder if it could also be proved that $u_n$ is never a square because $u_n-1<sqrtu_n<u_n-1+1$
$endgroup$
– J. W. Tanner
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The problem, indeed, is very cute.
We just need to check the condition $textrmmod 5$:
$$u_1 equiv 1 textrm mod 5$$
$$u_2 equiv 2 textrm mod 5$$
$$u_3 equiv 2 textrm mod 5$$
etc.
Therefore, one can show by induction that for any $n in mathbbN$:
$$u_n+1 equiv left[ n+1+sum_j=2^n (-1) right] textrm mod 5 equiv 2 textrm mod 5.$$
The latter is never true for squares, since squares are equal to either 1 or 4 mod 5.
To prove the generalization, we use the recurrence relation of type
$$u_n+1 = u_n^2+u_n+a.$$
Notice that $u_n+1>u_n^2$ since $a>0$.
Therefore, if $u_n+1$ is a perfect square, then $ageq u_n+1 > u_n$.
But $a>u_n$ is impossible since for any $n geq 3$:
$$u_n = u_n-1^2+u_n-1+a >a.$$
This gives us a contradiction, so $u_n$ is never a perfect square for any
$ngeq 3$.
answered 3 hours ago
JaneJane
1,258313
$endgroup$
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
3 hours ago
$begingroup$
That is very nice, and it is completely different than my solution, so I will accept it. Now, how about the generalization?
$endgroup$
– marty cohen
3 hours ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
3 hours ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
2 hours ago
add a comment |
$begingroup$
The recurrence can be rewritten as
begineqnarray*
u_n+1=u_n^2+u_n+1.
endeqnarray*
It is easy to show that
begineqnarray*
u_n^2 < u_n+1 <(u_n+1)^2.
endeqnarray*
Now observe that there is no whole numbers between $u_n$ & $u_n+1$.
answered 3 hours ago
Donald SplutterwitDonald Splutterwit
23.3k21446
$endgroup$
$begingroup$
This was the idea I expressed in my comment; I just didn't take the time to flesh it out; thank you for doing so
$endgroup$
– J. W. Tanner
2 hours ago
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%2f3225094%2fshow-solution-to-recurrence-is-never-a-square%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
$begingroup$
The problem, indeed, is very cute.
We just need to check the condition $textrmmod 5$:
$$u_1 equiv 1 textrm mod 5$$
$$u_2 equiv 2 textrm mod 5$$
$$u_3 equiv 2 textrm mod 5$$
etc.
Therefore, one can show by induction that for any $n in mathbbN$:
$$u_n+1 equiv left[ n+1+sum_j=2^n (-1) right] textrm mod 5 equiv 2 textrm mod 5.$$
The latter is never true for squares, since squares are equal to either 1 or 4 mod 5.
To prove the generalization, we use the recurrence relation of type
$$u_n+1 = u_n^2+u_n+a.$$
Notice that $u_n+1>u_n^2$ since $a>0$.
Therefore, if $u_n+1$ is a perfect square, then $ageq u_n+1 > u_n$.
But $a>u_n$ is impossible since for any $n geq 3$:
$$u_n = u_n-1^2+u_n-1+a >a.$$
This gives us a contradiction, so $u_n$ is never a perfect square for any
$ngeq 3$.
answered 3 hours ago
JaneJane
1,258313
$endgroup$
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
3 hours ago
$begingroup$
That is very nice, and it is completely different than my solution, so I will accept it. Now, how about the generalization?
$endgroup$
– marty cohen
3 hours ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
3 hours ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
2 hours ago
add a comment |
$begingroup$
The problem, indeed, is very cute.
We just need to check the condition $textrmmod 5$:
$$u_1 equiv 1 textrm mod 5$$
$$u_2 equiv 2 textrm mod 5$$
$$u_3 equiv 2 textrm mod 5$$
etc.
Therefore, one can show by induction that for any $n in mathbbN$:
$$u_n+1 equiv left[ n+1+sum_j=2^n (-1) right] textrm mod 5 equiv 2 textrm mod 5.$$
The latter is never true for squares, since squares are equal to either 1 or 4 mod 5.
To prove the generalization, we use the recurrence relation of type
$$u_n+1 = u_n^2+u_n+a.$$
Notice that $u_n+1>u_n^2$ since $a>0$.
Therefore, if $u_n+1$ is a perfect square, then $ageq u_n+1 > u_n$.
But $a>u_n$ is impossible since for any $n geq 3$:
$$u_n = u_n-1^2+u_n-1+a >a.$$
This gives us a contradiction, so $u_n$ is never a perfect square for any
$ngeq 3$.
answered 3 hours ago
JaneJane
1,258313
$endgroup$
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
3 hours ago
$begingroup$
That is very nice, and it is completely different than my solution, so I will accept it. Now, how about the generalization?
$endgroup$
– marty cohen
3 hours ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
3 hours ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
2 hours ago
add a comment |
$begingroup$
The problem, indeed, is very cute.
We just need to check the condition $textrmmod 5$:
$$u_1 equiv 1 textrm mod 5$$
$$u_2 equiv 2 textrm mod 5$$
$$u_3 equiv 2 textrm mod 5$$
etc.
Therefore, one can show by induction that for any $n in mathbbN$:
$$u_n+1 equiv left[ n+1+sum_j=2^n (-1) right] textrm mod 5 equiv 2 textrm mod 5.$$
The latter is never true for squares, since squares are equal to either 1 or 4 mod 5.
To prove the generalization, we use the recurrence relation of type
$$u_n+1 = u_n^2+u_n+a.$$
Notice that $u_n+1>u_n^2$ since $a>0$.
Therefore, if $u_n+1$ is a perfect square, then $ageq u_n+1 > u_n$.
But $a>u_n$ is impossible since for any $n geq 3$:
$$u_n = u_n-1^2+u_n-1+a >a.$$
This gives us a contradiction, so $u_n$ is never a perfect square for any
$ngeq 3$.
answered 3 hours ago
JaneJane
1,258313
$endgroup$
The problem, indeed, is very cute.
We just need to check the condition $textrmmod 5$:
$$u_1 equiv 1 textrm mod 5$$
$$u_2 equiv 2 textrm mod 5$$
$$u_3 equiv 2 textrm mod 5$$
etc.
Therefore, one can show by induction that for any $n in mathbbN$:
$$u_n+1 equiv left[ n+1+sum_j=2^n (-1) right] textrm mod 5 equiv 2 textrm mod 5.$$
The latter is never true for squares, since squares are equal to either 1 or 4 mod 5.
To prove the generalization, we use the recurrence relation of type
$$u_n+1 = u_n^2+u_n+a.$$
Notice that $u_n+1>u_n^2$ since $a>0$.
Therefore, if $u_n+1$ is a perfect square, then $ageq u_n+1 > u_n$.
But $a>u_n$ is impossible since for any $n geq 3$:
$$u_n = u_n-1^2+u_n-1+a >a.$$
This gives us a contradiction, so $u_n$ is never a perfect square for any
$ngeq 3$.
answered 3 hours ago
JaneJane
1,258313
edited 2 hours ago
answered 3 hours ago
JaneJane
1,258313
answered 3 hours ago
JaneJane
1,258313
answered 3 hours ago
JaneJane
1,258313
1,258313
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
3 hours ago
$begingroup$
That is very nice, and it is completely different than my solution, so I will accept it. Now, how about the generalization?
$endgroup$
– marty cohen
3 hours ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
3 hours ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
2 hours ago
add a comment |
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
3 hours ago
$begingroup$
That is very nice, and it is completely different than my solution, so I will accept it. Now, how about the generalization?
$endgroup$
– marty cohen
3 hours ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
3 hours ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
2 hours ago
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
3 hours ago
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
3 hours ago
$begingroup$
That is very nice, and it is completely different than my solution, so I will accept it. Now, how about the generalization?
$endgroup$
– marty cohen
3 hours ago
$begingroup$
That is very nice, and it is completely different than my solution, so I will accept it. Now, how about the generalization?
$endgroup$
– marty cohen
3 hours ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
3 hours ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
3 hours ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
2 hours ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
2 hours ago
add a comment |
$begingroup$
The recurrence can be rewritten as
begineqnarray*
u_n+1=u_n^2+u_n+1.
endeqnarray*
It is easy to show that
begineqnarray*
u_n^2 < u_n+1 <(u_n+1)^2.
endeqnarray*
Now observe that there is no whole numbers between $u_n$ & $u_n+1$.
answered 3 hours ago
Donald SplutterwitDonald Splutterwit
23.3k21446
$endgroup$
$begingroup$
This was the idea I expressed in my comment; I just didn't take the time to flesh it out; thank you for doing so
$endgroup$
– J. W. Tanner
2 hours ago
add a comment |
$begingroup$
The recurrence can be rewritten as
begineqnarray*
u_n+1=u_n^2+u_n+1.
endeqnarray*
It is easy to show that
begineqnarray*
u_n^2 < u_n+1 <(u_n+1)^2.
endeqnarray*
Now observe that there is no whole numbers between $u_n$ & $u_n+1$.
answered 3 hours ago
Donald SplutterwitDonald Splutterwit
23.3k21446
$endgroup$
$begingroup$
This was the idea I expressed in my comment; I just didn't take the time to flesh it out; thank you for doing so
$endgroup$
– J. W. Tanner
2 hours ago
add a comment |
$begingroup$
The recurrence can be rewritten as
begineqnarray*
u_n+1=u_n^2+u_n+1.
endeqnarray*
It is easy to show that
begineqnarray*
u_n^2 < u_n+1 <(u_n+1)^2.
endeqnarray*
Now observe that there is no whole numbers between $u_n$ & $u_n+1$.
answered 3 hours ago
Donald SplutterwitDonald Splutterwit
23.3k21446
$endgroup$
The recurrence can be rewritten as
begineqnarray*
u_n+1=u_n^2+u_n+1.
endeqnarray*
It is easy to show that
begineqnarray*
u_n^2 < u_n+1 <(u_n+1)^2.
endeqnarray*
Now observe that there is no whole numbers between $u_n$ & $u_n+1$.
answered 3 hours ago
Donald SplutterwitDonald Splutterwit
23.3k21446
answered 3 hours ago
Donald SplutterwitDonald Splutterwit
23.3k21446
answered 3 hours ago
Donald SplutterwitDonald Splutterwit
23.3k21446
answered 3 hours ago
Donald SplutterwitDonald Splutterwit
23.3k21446
23.3k21446
$begingroup$
This was the idea I expressed in my comment; I just didn't take the time to flesh it out; thank you for doing so
$endgroup$
– J. W. Tanner
2 hours ago
add a comment |
$begingroup$
This was the idea I expressed in my comment; I just didn't take the time to flesh it out; thank you for doing so
$endgroup$
– J. W. Tanner
2 hours ago
$begingroup$
This was the idea I expressed in my comment; I just didn't take the time to flesh it out; thank you for doing so
$endgroup$
– J. W. Tanner
2 hours ago
$begingroup$
This was the idea I expressed in my comment; I just didn't take the time to flesh it out; thank you for doing so
$endgroup$
– J. W. Tanner
2 hours ago
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%2f3225094%2fshow-solution-to-recurrence-is-never-a-square%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
1
$begingroup$
I see the answer from considering modulo $5$. I wonder if it could also be proved that $u_n$ is never a square because $u_n-1<sqrtu_n<u_n-1+1$
$endgroup$
– J. W. Tanner
3 hours ago