Find all integers $n$, $n \ge 1$, such that $n \cdot 2^{n+1}+1$ is a perfect square.
Problem
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Tags: induction, inequalities, number theory, prime factorization, number theory proposed
Aksenov239
20.06.2010 21:32
Let $n \cdot 2^{n + 1} + 1 = k^2;$ We have : $n \cdot 2^{n + 1} = (k - 1)(k + 1);$ We can see, that $k$ - odd! Because of that : 1); $k - 1 = 2^{n}z_1; k + 1 = 2z_2;$ $z_2 \leq n; \Rightarrow 2^{n} \leq 2n - 2$ - for $n \leq 3;$ 2); $k + 1 = 2^{n}z_1; k - 1 = 2z_2;$ $z_2 \leq n; \Rightarrow 2^{n} \leq 2n + 2$ - for $n \leq 3;$ We can see, that $n \leq 3;$ Because of that answer : $n = 3;$ That's all!
mahanmath
20.06.2010 21:51
$n \cdot 2^{n + 1} = (k - 1)(k + 1)$ So $n \cdot 2^{n-1 } = \frac{(k - 1)}{2} . \frac{(k + 1)}{2};$ And now $\frac{{k - 1}}{2}$ , $\frac{{k + 1}}{2}$ are consecutive and ....
Ahiles
20.06.2010 22:17
$n\cdot 2^{n+1}=(p-1)(p+1)$. Let $n=2^k a$, where $a$ is odd. Because $p-1$ and $p+1$ are two consecutive even numbers, one of them has exponent of $2$ equal to $1$ and other $n+k$. We have two cases $\bullet$ $p-1=2^{n+k}a'$ and $p+1=2a''$. Then \[2 a''=p+1>p-1=2^{n+k}a'\ge 2^{a''}\] But it can be easily checked that for $b\ge 1$, $2^{b-1}\ge b$, so we have a contradiction in this case. $\bullet$ $p-1=2a'$ and $p+1=2^{n+k}a''$. After substracting we get $2^{n+k-1}a''-a'=1$. But for $k\ge 1$ we have \[ 2^{n+k-1}a'' \ge 2^{2a'} \ge 2^{a'+1} a'> a'+1\] So, $k=0$. Equation transforms into \[2^{a'a''-1}a''=a'+1\] For $a'>3$ we have $2^{a'a''-1}a''\ge 2^{a'-1}>a'+1$. Checking $a'=1,2,3$ we find solution $n=3$.
Jason-Kountou
20.06.2010 23:28
Please excuse any mistakes made, I'm still rather new to this and I hope it is correct:
We have $ n2^{n+1}+1 = k^2 \Leftrightarrow n2^{n+1} = (k-1)(k+1)$. (1)
Splitting into three cases:
- $n \mid 1 \Leftrightarrow n =1$, which does not satisfy the above condition ($k^2 =5$)
- $ n \mid k-1 \Leftrightarrow k-1 = mn, m \in \mathbb{N}$. (2)
Plugging this in in the first equation gives us
$ 2^{n+1} = m(k+1)$. Since $2^{n+1}$ 's prime factorization composes only of 2's (thus no other primes), both m and k+1 must be a perfect power of 2 or k+1 is a power of 2 and m=1 (k+1 = 1 gives k =0, contradiction).
The case of m = 1 turns the first equation (with the help of (2)) to $2^{n+1} = n +2 $ which has an obvious solution $ n =0$ (which does not satisfy the conditions).
By induction, for $n =1$ we have $ 4 > 3$.
Assuming for a certain $k \in \mathbb{N} $ that $2^{k+1} > k +2$, we need to prove $ 2^{k+2} > k+3$.
But $ 2^{k+2} = 2^{k+1} \times 2 > 2(k+2) > k+3$, thus there are no other natural solutions.
Now let $ m = 2^p, k+1 = 2^q (p,q \neq 0, p+q = n+1)$ and (2) becomes $2^q - 2 = 2^pn \Leftrightarrow 2(2^{q-1}-1) = 2^pn$.
Obviously $2^p$ cannot divide $2^{q-1} - 1$, so $ 2^p \mid 2 \Leftrightarrow p =1, m = 2$. From there we also get $ q = n$ and the equation becomes $2^n - 2 = 2n$.
For $ n \geq 4$, by induction we can prove that $2^n > 2(n+1)$ (it follows a similiar pattern like the first induction).
Thus we check cases $ n =3,2$, of which $n=3$ works.
- $n \mid k+1$. As in case 2, we follow the same steps: $ k+1 = mn$ and (1) becomes $2^{n+1} = m(k-1)$.
If m = 1 then $ k = n-1$ and $ 2^{n+1} = n - 2$, which has no natural solutions (using the same induction steps again).
If k-1 = 1 then k = 2 and since $ n \mid 3$ we have $ n =3$, which works.
If $ m, k-1 \geq 2$, they must be perfect powers of 2 as said above.
Letting $ m = 2^p, k-1 = 2^q$ ( and $ p +q = n+1$) again turns (2) into $ 2(2^{q-1} +1) = 2^pn$ and as proven above, $ p =1, m = 2$ and $ q =n$.
Then the equation becomes $2^n = 2n -2$which by induction again is proven to have no natural solutions.
(for the record: Case $ n =1$ gives $ 2>0$ and case $ n =2$ gives $4>2$. Assuming the inequality holds for $k \in \mathbb{N}$, we need to prove $ 2^{k+1} > 2k$ but $ 2^{k+1} > 2(2k-2) > 2k, k \geq 3$.)
Thus I find $n =3$ to be the only solution.
.
Bugi
21.06.2010 00:16
The following fact is well known and easily proven with induction: $2^n\ge n^2$, for $n\ge 5$
Obvious idea: $n\cdot 2^{n+1}=(a-1)(a+1)$, also let $n=2^t\cdot u$, $u$ is odd.
There are 2 consecutive even numbers on the RHS, so one of them is not divisible by 4.
1) it's $a+1$
$a-1=2^{n+t}x,a+1=2y,xy=u$
$2^{n+t-1}x+1=y\le n$, which is impossible because $2^{n-1}\ge n$
2) it's $a-1$, assume $n\ge 4$
$a-1=2x,a+1=2^py,p=n+t$
Now, $x+1=2^{p-1}y$, so $x+1\ge 2^{p-1}$
$2^{\frac{n+1}{2}}\ge n+1\ge x+1\ge 2^{p-1}$, so $n+1\ge 2p-2$
We get that $p+1\le \frac{n+5}{2}$
$t+n+1\le\frac{n+5}{2}$, so $2t+n\le 3$, so $n\le 3$, which is a contradiction.
That means $n\le 3$, the only solution is for $n=3$
Lawasu
17.05.2012 13:50
Clearly $n\cdot 2^{n+1}+1$ is odd, so, if this number is a perfect square, then $n\cdot 2^{n+1}+1=(2x+1)^2$, where $x\in \Bbb{N}$, whence $n\cdot 2^{n-1}=x(x+1)$. The integers $x$ and $x+1$ are coprime, so one of them must be divisible by $2^{n-1}$, which means that the other must be at most $n$. This shows that $2^{n-1}\le n+1$. An easy induction shows that the above inequality is false for all $n\ge4$. A direct inspection confirms that the only convenient values in the case $n\le 3$ are $0$ and $3$.
viperstrike
31.10.2013 23:47
Really quick bounding solution since we need 2^(n-1)+1 <=n so n<=3. Inspection gives n=3 the only answer.
silouan
01.08.2015 18:21
See here: http://artofproblemsolving.com/community/c6h262470p1426044 The problem was already known before the competition
Kunihiko_Chikaya
02.08.2015 03:48
Aksenov239 wrote: Ahiles wrote: Find all integers $n$, $n \ge 1$, such that $n \cdot 2^{n+1}+1$ is a perfect square. Let $n \cdot 2^{n + 1} + 1 = k^2;$ We have : $n \cdot 2^{n + 1} = (k - 1)(k + 1);$ We can see, that $k$ - odd! Because of that : 1); $k - 1 = 2^{n}z_1; k + 1 = 2z_2;$ $z_2 \leq n; \Rightarrow 2^{n} \leq 2n - 2$ - for $n \leq 3;$ 2); $k + 1 = 2^{n}z_1; k - 1 = 2z_2;$ $z_2 \leq n; \Rightarrow 2^{n} \leq 2n + 2$ - for $n \leq 3;$ We can see, that $n \leq 3;$ Because of that answer : $n = 3;$ That's all! Aksenov239, Please don't quote the entire post before yours. From now on, we would mark the number of Alert, say, Alert 1, Alert 2, and so on. If some person could have 10 Alert, the member would be deleted the register in AoPS. Thanks in advance.
Ika789_Master
02.12.2017 14:49
n*2^n+1 + 1 is odd number so we have n*2^n+1 + 1 = (2m+1)^2 n*2^n+1 = (2m+1 - 1)(2m +1 +1) n*2^n+1=4k(k+1) n*2^n-1=k(k+1) From gcd(k, k+1) = 1 problem becomes trivial.
megarnie
30.06.2022 14:09
Let $n\cdot 2^{n+1} + 1 = k^2$. So $n\cdot 2^{n+1} = (k-1)(k+1)$. Note that this implies $k$ is odd. Case 1: $\nu_2(k-1) = n$ and $\nu_2(k+1) = 1$. Let $k+1=2c$. Then we have $c\le n$. So $2^n\le k-1 < k+1 = 2c \le 2n$, which implies $n\le 2$, both $n=1$ and $n=2$ don't work. Case 2: $\nu_2(k-1) = 1 $ and $\nu_2(k+1) = n$. Let $k-1 = 2c$. Then we have $c\le n$. So $2^n -2 \le (k+1)-2 = k-1 = 2c \le 2n$. so $2^n \le 2n+2\implies n\le 3$. Clearly only $\boxed{n=3}$ works.
rafigamath
17.03.2023 18:10
just use the fact that if $n$ and $n+2$ are two consecutive even integers then one of them is divisible by 2 at most,and the other one is divisible by $2^n$ ,where $n>=2$. You will be left with 4 cases and thats it.
huashiliao2020
19.04.2023 20:46
Let $n \cdot 2^{n + 1} + 1 = k^2; n*2^{n + 1}=(k - 1)(k + 1)$. Since k is odd, if $k=m*2^q+1$, then $k-1\equiv{0}$ mod 2^q, but $k+1\equiv{2}$ mod 2^q, implying k+1 (or k-1, since we did this WLOG) is a multiple of 2 but not one of 4. 1); $k - 1 = 2^{n}i; k + 1 = 2j$. This implies that $k+1\leq n$, or $2^n*j=k-1\leq 2^n\leq 2n-2$. We can either have the first one, or since we did this WLOG, we can similarly derive $2^n\leq 2n+2$. It is obvious that 2^n increases MUCH much faster than 2n+2, so we can get that $\boxed{n=3}$. Nice problem!
bin_sherlo
14.08.2023 22:08
Answer:$3$. $n.2^{n+1}=x^2-1=(x-1)(x+1)$ There are $4$ cases. $i)x-1=2n,x+1=2^n \implies n+1=2^{n-1} \implies n=3$ $ii)x-1=2,x+1=2^nn \implies 4=2^nn$ No solution $iii)x-1=2^nn,x+1=2 \implies $ No solution $iv)x-1=2^n,x+1=2n \implies 2^{n-1}=n-1\implies $ No solution