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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 1 hour 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 1 hour 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
1 hour 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 1 hour 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 1 hour ago
marty cohenmarty cohen
76.8k549130
asked 1 hour ago
marty cohenmarty cohen
76.8k549130
asked 1 hour ago
marty cohenmarty cohen
76.8k549130
asked 1 hour ago
marty cohenmarty cohen
76.8k549130
asked 1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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 1 hour ago
JaneJane
1,238313
$endgroup$
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
1 hour 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
1 hour ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
1 hour ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
41 mins 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 56 mins 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
39 mins ago
add a comment |
Your Answer
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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 1 hour ago
JaneJane
1,238313
$endgroup$
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
1 hour 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
1 hour ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
1 hour ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
41 mins 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 1 hour ago
JaneJane
1,238313
$endgroup$
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
1 hour 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
1 hour ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
1 hour ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
41 mins 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 1 hour ago
JaneJane
1,238313
$endgroup$
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 1 hour ago
JaneJane
1,238313
edited 41 mins ago
answered 1 hour ago
JaneJane
1,238313
answered 1 hour ago
JaneJane
1,238313
answered 1 hour ago
JaneJane
1,238313
1,238313
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
1 hour 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
1 hour ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
1 hour ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
41 mins ago
add a comment |
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
1 hour 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
1 hour ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
1 hour ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
41 mins ago
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
1 hour ago
$begingroup$
although, I need to think a bit more about the generalization
$endgroup$
– Jane
1 hour 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
1 hour 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
1 hour ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
1 hour ago
$begingroup$
thank you! i am thinking about the generalization right now.
$endgroup$
– Jane
1 hour ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
41 mins ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
41 mins 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 56 mins 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
39 mins 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 56 mins 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
39 mins 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 56 mins 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 56 mins ago
Donald SplutterwitDonald Splutterwit
23.3k21446
answered 56 mins ago
Donald SplutterwitDonald Splutterwit
23.3k21446
answered 56 mins ago
Donald SplutterwitDonald Splutterwit
23.3k21446
answered 56 mins 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
39 mins 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
39 mins 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
39 mins 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
39 mins ago
add a comment |
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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
1 hour ago