Hi! This week's set consists in three problems in algebra that seem nice. I'm using the word "seem" because I've been working on them and haven't been able to solve them. Anyway, enjoy the problems, and, if you find them nice and/or fun, please reshare the note, so more people will be able to see the problems and try to tackle these hard problems. Just as a reference, problem 1 was taken from the Colombian book "Un recorrido por el álgebra" and problems 2 and 3 were taken from the first Costa Rican IMO TST 2014.

Find all functions \(f:\mathbb{Z}^+\rightarrow\mathbb{R}\) such that \[f(n+m)+f(n-m)=f(3n)\] for every \(n,m\in{\mathbb{Z}^+}, n>m\).

Find all real solutions \((x_1,...,x_n)\) of the equation \[(1-x_1)^2+(x_1-x_2)^2+...+(x_{n-1}-x_n)^2+x_n^2=\dfrac{1}{n+1}\]

Let \(x, y\) be real numbers such that \(y>0\) and \(7x^2+3xy+3y^2=1\). Show that \[\dfrac{x^2+y^2}{y}\ge{\dfrac{1}{2}}\]

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TopNewestFor the second one:

Using the Cauchy-Schwarz Inequality we can write, \( \displaystyle \begin{array} \\ \bigg( (1-x_1)^2 + (x_1 - x_2 )^2 +\dotsb + (x_{n-1} - x_n)^2 + x_n^2 \bigg) \bigg( 1^2 + 1^2 + \dotsb + 1^2 + 1^2 \bigg) \geq & \\ \bigg( (1-x_1) + (x_1 - x_2 ) +\dotsb + (x_{n-1} - x_n)+ x_n \bigg)^2 & \\ \end{array} \)

\( \displaystyle \Rightarrow \Big( (1-x_1)^2 + (x_1 - x_2 )^2 +\dotsb + (x_{n-1} - x_n)^2 + x_n^2 \Big) (n+1) \geq ( 1)^2 \) \(\displaystyle \Rightarrow (1-x_1)^2 + (x_1 - x_2 )^2 +\dotsb + (x_{n-1} - x_n)^2 + x_n^2 \geq \frac{1}{n+1} \)

But it is given that the equality condition holds true. Therefore, from the condition of equality in Cauchy-Schwarz, we obtain,

\( \displaystyle \frac{1-x_1}{1} = \frac{x_1 - x_2}{1} = \dots = \frac{x_{n-1} - x_n}{1} = \frac{x_n}{1} \) \(\displaystyle \Rightarrow x_{n-1} = 2x_n \text{ , } x_{n-2} = 3x_n ,\dotsc, \text{ } x_2 = (n-1)x_n \text{ and } x_1 = nx_n \).

Also, \(\displaystyle 1 = x_1 + x_n \Rightarrow (n+1)x_n = 1 \Rightarrow x_n = \frac{1}{n+1} \).

Hence, \(\displaystyle x_i = \frac{n+1-i}{n+1} \text{ ; } i = 1,2,3, \dotsc , n-1, n \)

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May I know how you arrive at a conclusion for \(x_{i}\)?

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From third last line, \(x_{i} = (n+1-i)x_{n}\) and \(x_{n} = \frac{1}{n+1}\).

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For no.1 I have a weird start, this might helps.

Substituting \(m \rightarrow 1, 2, 3, ..., n-1\) we get

\(f(2n-i) + f(i) = f(3n) = f(2n-j) + f(j)\) for every \(i,j = 1, 2, ..., n-1\)

We see that if \(a+b = c+d\), then \(f(a)+f(b)=f(c)+f(d)\) for positive integers \(a,b,c,d\).

I have no idea to do anything right meow. I suck at this thing XD

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I know of a "crude" solution to 3, which only uses basic knowledge of conic sections. It has a nice interpretation that way.

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I'm afraid I don't know much about conic sections, but share the solution and I'll try to look up the stuff I don't understand :). Besides, it would be useful for other people who may know about conic sections.

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Show that the circle \( 2 x^2 + 2 y^2 - y = 0 \) is contained within the ellipse \( 7x^2 + 3xy + 3y^2 = 1 \). In fact, these conic sections are tangential at the point \( ( \frac{ 1}{5}, \frac{2}{5}) \).

Hence, it follows that for any point on the ellipse which satisfy \( y \geq 0 \), we have

\[ 2 x^2 + 2y^2 - y \geq 0 \Rightarrow \frac{ x^2+y^2} { y} \geq \frac{1}{2}. \]

The tangential point is the equality case.

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For question \(2\), does the solution only occur when \(n=0\) with \(x_{0}=1\)?

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Feel free to post ideas and solutions! :D

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