Find all integers a so that the sequence (Xn) satisfies the condition
$begingroup$
Let ($x_n$) be a sequence integers defined as such:
For all $n>1$ $x_{n+2} =2x_{n+1}x_{n} - x_{n+1} - x_n +1$ ,
$x_0=a$ , $x_1=2$
Find all integers a such that for all $n>1$ the number
$2x_{3n} -1$ is a square of a natural number.
Found this one in an old russian textbook for olympiad practice, couldn't solve it and the author didn't include the solution.
elementary-number-theory recurrence-relations contest-math recursion
edited Jan 15 at 16:44
amWhy
1
asked Jan 15 at 16:28
DoodDood
487
$endgroup$
add a comment |
$begingroup$
Let ($x_n$) be a sequence integers defined as such:
For all $n>1$ $x_{n+2} =2x_{n+1}x_{n} - x_{n+1} - x_n +1$ ,
$x_0=a$ , $x_1=2$
Find all integers a such that for all $n>1$ the number
$2x_{3n} -1$ is a square of a natural number.
Found this one in an old russian textbook for olympiad practice, couldn't solve it and the author didn't include the solution.
elementary-number-theory recurrence-relations contest-math recursion
edited Jan 15 at 16:44
amWhy
1
asked Jan 15 at 16:28
DoodDood
487
$endgroup$
add a comment |
$begingroup$
Let ($x_n$) be a sequence integers defined as such:
For all $n>1$ $x_{n+2} =2x_{n+1}x_{n} - x_{n+1} - x_n +1$ ,
$x_0=a$ , $x_1=2$
Find all integers a such that for all $n>1$ the number
$2x_{3n} -1$ is a square of a natural number.
Found this one in an old russian textbook for olympiad practice, couldn't solve it and the author didn't include the solution.
elementary-number-theory recurrence-relations contest-math recursion
edited Jan 15 at 16:44
amWhy
1
asked Jan 15 at 16:28
DoodDood
487
$endgroup$
Let ($x_n$) be a sequence integers defined as such:
For all $n>1$ $x_{n+2} =2x_{n+1}x_{n} - x_{n+1} - x_n +1$ ,
$x_0=a$ , $x_1=2$
Find all integers a such that for all $n>1$ the number
$2x_{3n} -1$ is a square of a natural number.
Found this one in an old russian textbook for olympiad practice, couldn't solve it and the author didn't include the solution.
elementary-number-theory recurrence-relations contest-math recursion
elementary-number-theory recurrence-relations contest-math recursion
edited Jan 15 at 16:44
amWhy
1
asked Jan 15 at 16:28
DoodDood
487
edited Jan 15 at 16:44
amWhy
1
asked Jan 15 at 16:28
DoodDood
487
edited Jan 15 at 16:44
amWhy
1
edited Jan 15 at 16:44
amWhy
1
edited Jan 15 at 16:44
amWhy
1
1
asked Jan 15 at 16:28
DoodDood
487
asked Jan 15 at 16:28
DoodDood
487
asked Jan 15 at 16:28
DoodDood
487
487
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $y_n=2x_n-1$. Then $y_0=2a-1$, $y_1=5$, and
$$y_{n+2}= y_{n+1}y_n.$$
The power $v_5(y_n)$ with which $5$ occurs in $y_n$ is then governed by a Fibonacci-like recursion
$$v_5(y_{n+2})=v_5(y_{n+1})v_5(y_{n}) $$
where we only know $v_5(y_1)=1$ exactly and $v_5(y_0)=v_5(2a-1)$ is some finite integer.
We conclude that
$$v_5(y_n)=v_5(2a-1)cdot F_{n-1} +F_n.tag1$$
For all other primes $pne 5$, we find
$$v_p(y_n)=v_p(2a-1)cdot F_{n-1}.tag2$$
We want that both $(1)$ and $(2)$ are even whenever $n$ is a multiple of $3$ and $>3$.
One readily verifies, that for such $n$, $F_n$ is even and $F_{n-1}$ is odd. We conclude that the desired property holds iff $v_(2a-1)$ is even for all $p$, i.e., $$2a-1text{ is a perfect square},$$
or in other words, $$a=frac{(2k-1)^2+1}2=2k(k-1)+1$$ for some $kinBbb N$.
answered Jan 15 at 17:21
Hagen von EitzenHagen von Eitzen
279k23270501
$endgroup$
$begingroup$
Thank you for that contribution.
$endgroup$
– Dood
Jan 15 at 17:33
add a comment |
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1 Answer
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$begingroup$
Let $y_n=2x_n-1$. Then $y_0=2a-1$, $y_1=5$, and
$$y_{n+2}= y_{n+1}y_n.$$
The power $v_5(y_n)$ with which $5$ occurs in $y_n$ is then governed by a Fibonacci-like recursion
$$v_5(y_{n+2})=v_5(y_{n+1})v_5(y_{n}) $$
where we only know $v_5(y_1)=1$ exactly and $v_5(y_0)=v_5(2a-1)$ is some finite integer.
We conclude that
$$v_5(y_n)=v_5(2a-1)cdot F_{n-1} +F_n.tag1$$
For all other primes $pne 5$, we find
$$v_p(y_n)=v_p(2a-1)cdot F_{n-1}.tag2$$
We want that both $(1)$ and $(2)$ are even whenever $n$ is a multiple of $3$ and $>3$.
One readily verifies, that for such $n$, $F_n$ is even and $F_{n-1}$ is odd. We conclude that the desired property holds iff $v_(2a-1)$ is even for all $p$, i.e., $$2a-1text{ is a perfect square},$$
or in other words, $$a=frac{(2k-1)^2+1}2=2k(k-1)+1$$ for some $kinBbb N$.
answered Jan 15 at 17:21
Hagen von EitzenHagen von Eitzen
279k23270501
$endgroup$
$begingroup$
Thank you for that contribution.
$endgroup$
– Dood
Jan 15 at 17:33
add a comment |
$begingroup$
Let $y_n=2x_n-1$. Then $y_0=2a-1$, $y_1=5$, and
$$y_{n+2}= y_{n+1}y_n.$$
The power $v_5(y_n)$ with which $5$ occurs in $y_n$ is then governed by a Fibonacci-like recursion
$$v_5(y_{n+2})=v_5(y_{n+1})v_5(y_{n}) $$
where we only know $v_5(y_1)=1$ exactly and $v_5(y_0)=v_5(2a-1)$ is some finite integer.
We conclude that
$$v_5(y_n)=v_5(2a-1)cdot F_{n-1} +F_n.tag1$$
For all other primes $pne 5$, we find
$$v_p(y_n)=v_p(2a-1)cdot F_{n-1}.tag2$$
We want that both $(1)$ and $(2)$ are even whenever $n$ is a multiple of $3$ and $>3$.
One readily verifies, that for such $n$, $F_n$ is even and $F_{n-1}$ is odd. We conclude that the desired property holds iff $v_(2a-1)$ is even for all $p$, i.e., $$2a-1text{ is a perfect square},$$
or in other words, $$a=frac{(2k-1)^2+1}2=2k(k-1)+1$$ for some $kinBbb N$.
answered Jan 15 at 17:21
Hagen von EitzenHagen von Eitzen
279k23270501
$endgroup$
$begingroup$
Thank you for that contribution.
$endgroup$
– Dood
Jan 15 at 17:33
add a comment |
$begingroup$
Let $y_n=2x_n-1$. Then $y_0=2a-1$, $y_1=5$, and
$$y_{n+2}= y_{n+1}y_n.$$
The power $v_5(y_n)$ with which $5$ occurs in $y_n$ is then governed by a Fibonacci-like recursion
$$v_5(y_{n+2})=v_5(y_{n+1})v_5(y_{n}) $$
where we only know $v_5(y_1)=1$ exactly and $v_5(y_0)=v_5(2a-1)$ is some finite integer.
We conclude that
$$v_5(y_n)=v_5(2a-1)cdot F_{n-1} +F_n.tag1$$
For all other primes $pne 5$, we find
$$v_p(y_n)=v_p(2a-1)cdot F_{n-1}.tag2$$
We want that both $(1)$ and $(2)$ are even whenever $n$ is a multiple of $3$ and $>3$.
One readily verifies, that for such $n$, $F_n$ is even and $F_{n-1}$ is odd. We conclude that the desired property holds iff $v_(2a-1)$ is even for all $p$, i.e., $$2a-1text{ is a perfect square},$$
or in other words, $$a=frac{(2k-1)^2+1}2=2k(k-1)+1$$ for some $kinBbb N$.
answered Jan 15 at 17:21
Hagen von EitzenHagen von Eitzen
279k23270501
$endgroup$
Let $y_n=2x_n-1$. Then $y_0=2a-1$, $y_1=5$, and
$$y_{n+2}= y_{n+1}y_n.$$
The power $v_5(y_n)$ with which $5$ occurs in $y_n$ is then governed by a Fibonacci-like recursion
$$v_5(y_{n+2})=v_5(y_{n+1})v_5(y_{n}) $$
where we only know $v_5(y_1)=1$ exactly and $v_5(y_0)=v_5(2a-1)$ is some finite integer.
We conclude that
$$v_5(y_n)=v_5(2a-1)cdot F_{n-1} +F_n.tag1$$
For all other primes $pne 5$, we find
$$v_p(y_n)=v_p(2a-1)cdot F_{n-1}.tag2$$
We want that both $(1)$ and $(2)$ are even whenever $n$ is a multiple of $3$ and $>3$.
One readily verifies, that for such $n$, $F_n$ is even and $F_{n-1}$ is odd. We conclude that the desired property holds iff $v_(2a-1)$ is even for all $p$, i.e., $$2a-1text{ is a perfect square},$$
or in other words, $$a=frac{(2k-1)^2+1}2=2k(k-1)+1$$ for some $kinBbb N$.
answered Jan 15 at 17:21
Hagen von EitzenHagen von Eitzen
279k23270501
answered Jan 15 at 17:21
Hagen von EitzenHagen von Eitzen
279k23270501
answered Jan 15 at 17:21
Hagen von EitzenHagen von Eitzen
279k23270501
answered Jan 15 at 17:21
Hagen von EitzenHagen von Eitzen
279k23270501
279k23270501
$begingroup$
Thank you for that contribution.
$endgroup$
– Dood
Jan 15 at 17:33
add a comment |
$begingroup$
Thank you for that contribution.
$endgroup$
– Dood
Jan 15 at 17:33
$begingroup$
Thank you for that contribution.
$endgroup$
– Dood
Jan 15 at 17:33
$begingroup$
Thank you for that contribution.
$endgroup$
– Dood
Jan 15 at 17:33
add a comment |
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