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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
1,258313
$endgroup$
$begingroup$
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
1 hour 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 1 hour 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
1 hour 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
1,258313
$endgroup$
$begingroup$
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
1 hour 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
1,258313
$endgroup$
$begingroup$
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
1 hour 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
1,258313
$endgroup$
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 1 hour ago
JaneJane
1,258313
edited 1 hour ago
answered 1 hour ago
JaneJane
1,258313
answered 1 hour ago
JaneJane
1,258313
answered 1 hour ago
JaneJane
1,258313
1,258313
$begingroup$
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
1 hour 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
1 hour 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
1 hour ago
$begingroup$
I added the proof for the generalization by editing this post.
$endgroup$
– Jane
1 hour 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 1 hour 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
1 hour 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 1 hour 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
1 hour 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 1 hour 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 1 hour ago
Donald SplutterwitDonald Splutterwit
23.3k21446
answered 1 hour ago
Donald SplutterwitDonald Splutterwit
23.3k21446
answered 1 hour ago
Donald SplutterwitDonald Splutterwit
23.3k21446
answered 1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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