Find all positive integers $n$ for which there exist positive integers $x_1, x_2, \dots, x_n$ such that $$ \frac{1}{x_1^2}+\frac{2}{x_2^2}+\frac{2^2}{x_3^2}+\cdots +\frac{2^{n-1}}{x_n^2}=1.$$
Problem
Source: 2020 MEMO I-4
Tags: algebra, memo, MEMO 2020, equation, induction
danepale
30.08.2020 19:58
Although not really number theory. (Is the I-2 compensating for this, and I-1 compensating for I-2?)
Tintarn
30.08.2020 20:16
All $n$ except $n=2$.
Claim: If $n$ has a solution, then $n+2$ has a solution.
Proof of claim: Let $(x_1,\dots,x_n)$ be a solution for $n$.
Then just check that $(x_1,\dots,x_{n-1},2x_n,2x_n,4x_n)$ is a solution for $n+2$.
Now note that $1$ is a solution for $n=1$ and $(3,3,3,6)$ is a solution for $n=4$ so that all $n \ne 2$ work.
Finally for $n=2$ the equation is equivalent to $(x_1^2-1)(x_2^2-2)=2$ which is easily seen to not have solution.
Kimchiks926
31.08.2020 16:03
Solved together with @blastoor.
We begin with case, when $n$ is odd. We claim that $x_1=x_2 =\ldots =x_{n-1} = \sqrt{2^{n-1}}; x_n = 2^{n-1}$ works. Note that:
\begin{align*}
\frac{1}{x_1^2} + \frac{2}{x_2^2} + \ldots + \frac{2^{n-2}}{x_{n-1}^2} + \frac{2^{n-1}}{x_n^2} = \\
=\frac{1+2+\ldots + 2^{n-2}}{2^{n-1}} + \frac{1}{2^{n-1}} = \\
=\frac{2^{n-1}-1+1}{2^{n-1}} = \\
=1
\end{align*}as desired.
Now we consider the case when $n$ is even integer. We can easily check that for $n=2$ there is no such positive integers $x_1,x_2$. Now we construct desired tuple for all even integer greater than $2$. Assume that $n = 2k+2$. We take $x_n = 3 \cdot 2^{k-1} $. Then we need the that the sum of remaining $ n -1$ numbers is $\frac{1}{9}$. Since $n-1$ is odd, we have proved that we can find such $x_1, \ldots , x_{n-1}$ that:
$$ \frac{1}{x_1^2} + \ldots + \frac{2^{n-2}}{x_{n-1}^2} = 1 $$If we multiply each number $x_1, \ldots, x_{n-1}$ by $3$ we get that sum of remaining $n-1$ summands will be $\frac{1}{9}$ as desired.
Mathematicsislovely
21.09.2020 20:31
The answer is all natural number $n\ge 3$ and $n=1$.There is nothing to prove when $n=1$. Claim:There does not exists $x,y\in \mathbb N$ such that $\frac{1}{x^2}+\frac{2}{y^2}=1$ proof.The equation is same as $(y^2-1)(x^2-2)=2$ which does not give any solution in $\mathbb N^2$. $\blacksquare$ Claim:For all odd positive integer $n$ there exists $n$ integers $x_1, x_2, \dots, x_n$ satisfying the problems property. proof.We will proceed by induction on $n$.For $n=1$ there is nothing to prove.For $n=3$ observe that $\frac{1}{4}+\frac{2}{4}+\frac{2^2}{16}=1$. Suppose for some odd $n$ we have $\frac{1}{x_1^2}+\frac{2}{x_2^2}+\frac{2^2}{x_3^2}+\cdots +\frac{2^{n-1}}{x_n^2}=1$.Then observe that $\frac{1}{4}[\frac{1}{x_1^2}+\frac{2}{x_2^2}+\cdots +\frac{2^{n-1}}{x_n^2}]+\frac{2^n}{2^{n+1}}+\frac{2^{n+1}}{2^{n+3}}=1$.And $2^{n+1},2^{n+2}$ are both perfect squares.Hence all odd $n$ works. $\blacksquare$ Claim:For all even integer $n\ge 4$ there are $x_1, x_2, \dots, x_n$ satisfying the problems condition. proof.For $n=4$ we have $\frac{1}{9}+\frac{2}{9}+\frac{2^2}{9}+\frac{2^3}{9\times 2^2}=1$. Suppose for some even $n$ we have $\frac{1}{x_1^2}+\frac{2}{x_2^2}+\frac{2^2}{x_3^2}+\cdots +\frac{2^{n-1}}{x_n^2}=1$.Then observe that $\frac{1}{4}[\frac{1}{x_1^2}+\frac{2}{x_2^2}+\cdots +\frac{2^{n-1}}{x_n^2}]+\frac{2^n}{2^{n+2}}+\frac{2^{n+1}}{2^{n+2}}=1$.And $2^{n+2}$ is perfect square.Hence all even $n$ works.$\blacksquare$
508669
12.03.2021 11:11
XbenX wrote: Find all positive integers $n$ for which there exist positive integers $x_1, x_2, \dots, x_n$ such that $$ \frac{1}{x_1^2}+\frac{2}{x_2^2}+\frac{2^2}{x_3^2}+\cdots +\frac{2^{n-1}}{x_n^2}=1.$$ We claim that all positive integers $n$ except $n = 2$ work. If $n = 2$, then we get that $\dfrac{1}{x_1^2} + \dfrac{2}{x_2^2} = 1 \implies 2x_1^2 + x_2^2 = x_1^2x_2^2 \implies 2x_1^2 = x_2^2(x_1^2-1)$, but $\text{gcd}(x_1^2, x_1^2-1) = 1$ implies that $2 = x_1^2-1$, a contradiction so no solution for $n = 2$. For $n = 1$, $x_1 = 1$ works, for $n = 3$ $x_1 = 2, x_2 = 2, x_3 = 4$ works and for $n = 4$, we have $x_1 = x_2 = x_3 = 3$ and $x_4 = 6$ which works. Now, let us say for some $n \neq 3$, we have a solution for $x_1, x_2, x_3 \dots x_n$. For $n_1 = n + 2$, choose the solution $y_i = x_i$ for all $i \leq n-1$, $y_n = 2x_n, y_{n+1} = 2x_n$ and $y_{n+2} = 4x_n$ and we see that $$\sum\limits_{j=1}^{n+2} \dfrac{j}{y_j^2} = \left ( 1 - \dfrac{3\cdot 2^{n-1}}{4x_n^2}\right ) + \dfrac{2^n}{4x_n^2} + \dfrac{2^{n+1}}{16x_n^2} = 1 - \dfrac{3 \cdot 2^{n+1} + 2^{n+2} + 2^{n+1}}{16x_n^2} = 1$$as desired.
lazizbek42
09.01.2022 18:37
Induction: for $n$ $$x_1,x_2,...,x_n$$for $n+2$ $$2,2,4x_1,...,4x_n$$Very nice problem